Find all possible values $sqrti+sqrt-i$ The Next CEO of Stack OverflowSimplifying $|a+b|^2 + |a-b|^2$Find all complex number $zinBbbC$ such that $vert zvert=vert z^-1vert=vert z-1vert$Find all values of $sqrt[4]-1+i$Find all pairs of values $a$ and $b$ that satisfy $(a+bi)^2 = 48 + 14i$Considering the complex number $z = m+i$ for which values of $m$ do we have $ left|overlinez+frac2zright| ge 1 $Find all solutions to $|z+sqrtz^2-1|=1$Finding all possible solutonsHow to find all values of $z$ such that $z^3=-8i$Find all values of$(1+i)^i=?$Find $z$ if given $|z|=sqrt3$ and $Re(z)=fracsqrt22$

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Find all possible values $sqrti+sqrt-i$



The Next CEO of Stack OverflowSimplifying $|a+b|^2 + |a-b|^2$Find all complex number $zinBbbC$ such that $vert zvert=vert z^-1vert=vert z-1vert$Find all values of $sqrt[4]-1+i$Find all pairs of values $a$ and $b$ that satisfy $(a+bi)^2 = 48 + 14i$Considering the complex number $z = m+i$ for which values of $m$ do we have $ left|overlinez+frac2zright| ge 1 $Find all solutions to $|z+sqrtz^2-1|=1$Finding all possible solutonsHow to find all values of $z$ such that $z^3=-8i$Find all values of$(1+i)^i=?$Find $z$ if given $|z|=sqrt3$ and $Re(z)=fracsqrt22$










0












$begingroup$



Find all possible values of $sqrti + sqrt-i$. Here $i= sqrt-1$.




My solution which didn't work.



begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray
begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+2abi &=& -1-i-2i\ &=& -1-3iendeqnarray
begincasesa^2-b^2=-1\2ab=-3endcases
begincasesa^2-b^2=-1\4a^2b^2=9endcases
begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases
begineqnarray2a^2 &=& sqrt10-1\a^2=fracsqrt10-12\ a &=& sqrtfracsqrt10-12\b &=& sqrtfracsqrt10+12endeqnarray










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$endgroup$







  • 1




    $begingroup$
    Please do not use pictures.
    $endgroup$
    – Dietrich Burde
    Mar 20 at 10:26















0












$begingroup$



Find all possible values of $sqrti + sqrt-i$. Here $i= sqrt-1$.




My solution which didn't work.



begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray
begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+2abi &=& -1-i-2i\ &=& -1-3iendeqnarray
begincasesa^2-b^2=-1\2ab=-3endcases
begincasesa^2-b^2=-1\4a^2b^2=9endcases
begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases
begineqnarray2a^2 &=& sqrt10-1\a^2=fracsqrt10-12\ a &=& sqrtfracsqrt10-12\b &=& sqrtfracsqrt10+12endeqnarray










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Please do not use pictures.
    $endgroup$
    – Dietrich Burde
    Mar 20 at 10:26













0












0








0





$begingroup$



Find all possible values of $sqrti + sqrt-i$. Here $i= sqrt-1$.




My solution which didn't work.



begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray
begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+2abi &=& -1-i-2i\ &=& -1-3iendeqnarray
begincasesa^2-b^2=-1\2ab=-3endcases
begincasesa^2-b^2=-1\4a^2b^2=9endcases
begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases
begineqnarray2a^2 &=& sqrt10-1\a^2=fracsqrt10-12\ a &=& sqrtfracsqrt10-12\b &=& sqrtfracsqrt10+12endeqnarray










share|cite|improve this question











$endgroup$





Find all possible values of $sqrti + sqrt-i$. Here $i= sqrt-1$.




My solution which didn't work.



begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray
begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+2abi &=& -1-i-2i\ &=& -1-3iendeqnarray
begincasesa^2-b^2=-1\2ab=-3endcases
begincasesa^2-b^2=-1\4a^2b^2=9endcases
begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases
begineqnarray2a^2 &=& sqrt10-1\a^2=fracsqrt10-12\ a &=& sqrtfracsqrt10-12\b &=& sqrtfracsqrt10+12endeqnarray







complex-numbers






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share|cite|improve this question













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share|cite|improve this question








edited Mar 20 at 11:57







Aryan Pandey

















asked Mar 20 at 10:24









Aryan PandeyAryan Pandey

13




13







  • 1




    $begingroup$
    Please do not use pictures.
    $endgroup$
    – Dietrich Burde
    Mar 20 at 10:26












  • 1




    $begingroup$
    Please do not use pictures.
    $endgroup$
    – Dietrich Burde
    Mar 20 at 10:26







1




1




$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Mar 20 at 10:26




$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Mar 20 at 10:26










4 Answers
4






active

oldest

votes


















0












$begingroup$

I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.



The misconception comes from the fact that we're used to thinking of $sqrtalpha$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=sqrt-1,$" unfortunately.



In particular, we must keep this in mind for any complex number $z$: $$sqrtz^2=pm z.tag$star$$$




begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray




We've got to be a little careful trying to use the property $$sqrtalphabeta=sqrtalphasqrtbeta$$ when $alpha$ and $beta$ aren't nonnegative reals, because of $(star).$ We should instead have $$a+ib=sqrtipmsqrtii.$$




begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+abi &=& -1-i-2i\ &=& -1-3iendeqnarray




I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=pmsqrt2,$ giving us $pmsqrt2i$ as two of our possible solutions. Considering also the case $a+ib=sqrti-sqrtii,$ we obtain instead $b=0$ and $a=pmsqrt 2,$ which gives us the other two solutions: $pmsqrt2$!



However, even if this had been correct, some more things went wrong on the way, as well.




begincasesa^2-b^2=-1\2ab=-3endcases
begincasesa^2-b^2=-1\4a^2b^2=9endcases
begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases




Fine so far, but only because we can't have $a^2+b^2=-sqrt10$ when $a,b$ are real!




begineqnarray2a^2 &=& sqrt10-1\a^2 &=&fracsqrt10-12\ a &=& sqrtfracsqrt10-12endeqnarray




Here, again, we should have used $(star)$ to instead have obtained $$a=pmsqrtfracsqrt10-12,$$ and similarly, we'd have had $$b=pmsqrtfracsqrt10+12.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Great , thank you very much, because of you guys , I really like math a lot.
    $endgroup$
    – Aryan Pandey
    Mar 20 at 12:04










  • $begingroup$
    You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
    $endgroup$
    – Cameron Buie
    Mar 20 at 12:36


















1












$begingroup$

Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    How you got idea , to use this expression?
    $endgroup$
    – Aryan Pandey
    Mar 20 at 10:39


















0












$begingroup$

Hint



Express $$i=e^ipiover2\-i=e^i3piover2$$then $$x_1=r_1e^itheta_1 , x_1^2=i\x_2=r_2e^itheta_2 , x^2_2=-i$$what are the possible values of $x_1+x_2$?






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    Hint:



    If $(a+bi)^2=alpha+beta i$, then:
    $$a=pmsqrtfracalphapmsqrtalpha^2+beta^22, b=pmsqrtfrac-alphapmsqrtalpha^2+beta^22$$



    Can you use this to find $sqrt i$ and $sqrt-i$ in the form $a+bi$?



    If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Got it , thanks
      $endgroup$
      – Aryan Pandey
      Mar 20 at 11:00











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.



    The misconception comes from the fact that we're used to thinking of $sqrtalpha$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=sqrt-1,$" unfortunately.



    In particular, we must keep this in mind for any complex number $z$: $$sqrtz^2=pm z.tag$star$$$




    begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray




    We've got to be a little careful trying to use the property $$sqrtalphabeta=sqrtalphasqrtbeta$$ when $alpha$ and $beta$ aren't nonnegative reals, because of $(star).$ We should instead have $$a+ib=sqrtipmsqrtii.$$




    begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+abi &=& -1-i-2i\ &=& -1-3iendeqnarray




    I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=pmsqrt2,$ giving us $pmsqrt2i$ as two of our possible solutions. Considering also the case $a+ib=sqrti-sqrtii,$ we obtain instead $b=0$ and $a=pmsqrt 2,$ which gives us the other two solutions: $pmsqrt2$!



    However, even if this had been correct, some more things went wrong on the way, as well.




    begincasesa^2-b^2=-1\2ab=-3endcases
    begincasesa^2-b^2=-1\4a^2b^2=9endcases
    begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
    begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases




    Fine so far, but only because we can't have $a^2+b^2=-sqrt10$ when $a,b$ are real!




    begineqnarray2a^2 &=& sqrt10-1\a^2 &=&fracsqrt10-12\ a &=& sqrtfracsqrt10-12endeqnarray




    Here, again, we should have used $(star)$ to instead have obtained $$a=pmsqrtfracsqrt10-12,$$ and similarly, we'd have had $$b=pmsqrtfracsqrt10+12.$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Great , thank you very much, because of you guys , I really like math a lot.
      $endgroup$
      – Aryan Pandey
      Mar 20 at 12:04










    • $begingroup$
      You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
      $endgroup$
      – Cameron Buie
      Mar 20 at 12:36















    0












    $begingroup$

    I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.



    The misconception comes from the fact that we're used to thinking of $sqrtalpha$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=sqrt-1,$" unfortunately.



    In particular, we must keep this in mind for any complex number $z$: $$sqrtz^2=pm z.tag$star$$$




    begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray




    We've got to be a little careful trying to use the property $$sqrtalphabeta=sqrtalphasqrtbeta$$ when $alpha$ and $beta$ aren't nonnegative reals, because of $(star).$ We should instead have $$a+ib=sqrtipmsqrtii.$$




    begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+abi &=& -1-i-2i\ &=& -1-3iendeqnarray




    I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=pmsqrt2,$ giving us $pmsqrt2i$ as two of our possible solutions. Considering also the case $a+ib=sqrti-sqrtii,$ we obtain instead $b=0$ and $a=pmsqrt 2,$ which gives us the other two solutions: $pmsqrt2$!



    However, even if this had been correct, some more things went wrong on the way, as well.




    begincasesa^2-b^2=-1\2ab=-3endcases
    begincasesa^2-b^2=-1\4a^2b^2=9endcases
    begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
    begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases




    Fine so far, but only because we can't have $a^2+b^2=-sqrt10$ when $a,b$ are real!




    begineqnarray2a^2 &=& sqrt10-1\a^2 &=&fracsqrt10-12\ a &=& sqrtfracsqrt10-12endeqnarray




    Here, again, we should have used $(star)$ to instead have obtained $$a=pmsqrtfracsqrt10-12,$$ and similarly, we'd have had $$b=pmsqrtfracsqrt10+12.$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Great , thank you very much, because of you guys , I really like math a lot.
      $endgroup$
      – Aryan Pandey
      Mar 20 at 12:04










    • $begingroup$
      You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
      $endgroup$
      – Cameron Buie
      Mar 20 at 12:36













    0












    0








    0





    $begingroup$

    I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.



    The misconception comes from the fact that we're used to thinking of $sqrtalpha$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=sqrt-1,$" unfortunately.



    In particular, we must keep this in mind for any complex number $z$: $$sqrtz^2=pm z.tag$star$$$




    begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray




    We've got to be a little careful trying to use the property $$sqrtalphabeta=sqrtalphasqrtbeta$$ when $alpha$ and $beta$ aren't nonnegative reals, because of $(star).$ We should instead have $$a+ib=sqrtipmsqrtii.$$




    begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+abi &=& -1-i-2i\ &=& -1-3iendeqnarray




    I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=pmsqrt2,$ giving us $pmsqrt2i$ as two of our possible solutions. Considering also the case $a+ib=sqrti-sqrtii,$ we obtain instead $b=0$ and $a=pmsqrt 2,$ which gives us the other two solutions: $pmsqrt2$!



    However, even if this had been correct, some more things went wrong on the way, as well.




    begincasesa^2-b^2=-1\2ab=-3endcases
    begincasesa^2-b^2=-1\4a^2b^2=9endcases
    begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
    begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases




    Fine so far, but only because we can't have $a^2+b^2=-sqrt10$ when $a,b$ are real!




    begineqnarray2a^2 &=& sqrt10-1\a^2 &=&fracsqrt10-12\ a &=& sqrtfracsqrt10-12endeqnarray




    Here, again, we should have used $(star)$ to instead have obtained $$a=pmsqrtfracsqrt10-12,$$ and similarly, we'd have had $$b=pmsqrtfracsqrt10+12.$$






    share|cite|improve this answer











    $endgroup$



    I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.



    The misconception comes from the fact that we're used to thinking of $sqrtalpha$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=sqrt-1,$" unfortunately.



    In particular, we must keep this in mind for any complex number $z$: $$sqrtz^2=pm z.tag$star$$$




    begineqnarraya+ib &=& sqrti+sqrt-i\ &=& sqrti+sqrti^2i\ &=& sqrti+sqrtiiendeqnarray




    We've got to be a little careful trying to use the property $$sqrtalphabeta=sqrtalphasqrtbeta$$ when $alpha$ and $beta$ aren't nonnegative reals, because of $(star).$ We should instead have $$a+ib=sqrtipmsqrtii.$$




    begineqnarray(a+ib)^2 &=& left(sqrti+sqrtiiright)^2\a^2-b^2+abi &=& -1-i-2i\ &=& -1-3iendeqnarray




    I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=pmsqrt2,$ giving us $pmsqrt2i$ as two of our possible solutions. Considering also the case $a+ib=sqrti-sqrtii,$ we obtain instead $b=0$ and $a=pmsqrt 2,$ which gives us the other two solutions: $pmsqrt2$!



    However, even if this had been correct, some more things went wrong on the way, as well.




    begincasesa^2-b^2=-1\2ab=-3endcases
    begincasesa^2-b^2=-1\4a^2b^2=9endcases
    begineqnarrayleft(a^2+b^2right)^2 &=& left(a^2-b^2right)^2+4a^2b^2\ &=& 1+9\ &=& 10endeqnarray
    begincasesa^2-b^2=-1\a^2+b^2=sqrt10endcases




    Fine so far, but only because we can't have $a^2+b^2=-sqrt10$ when $a,b$ are real!




    begineqnarray2a^2 &=& sqrt10-1\a^2 &=&fracsqrt10-12\ a &=& sqrtfracsqrt10-12endeqnarray




    Here, again, we should have used $(star)$ to instead have obtained $$a=pmsqrtfracsqrt10-12,$$ and similarly, we'd have had $$b=pmsqrtfracsqrt10+12.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 20 at 12:36

























    answered Mar 20 at 12:00









    Cameron BuieCameron Buie

    86.3k773161




    86.3k773161











    • $begingroup$
      Great , thank you very much, because of you guys , I really like math a lot.
      $endgroup$
      – Aryan Pandey
      Mar 20 at 12:04










    • $begingroup$
      You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
      $endgroup$
      – Cameron Buie
      Mar 20 at 12:36
















    • $begingroup$
      Great , thank you very much, because of you guys , I really like math a lot.
      $endgroup$
      – Aryan Pandey
      Mar 20 at 12:04










    • $begingroup$
      You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
      $endgroup$
      – Cameron Buie
      Mar 20 at 12:36















    $begingroup$
    Great , thank you very much, because of you guys , I really like math a lot.
    $endgroup$
    – Aryan Pandey
    Mar 20 at 12:04




    $begingroup$
    Great , thank you very much, because of you guys , I really like math a lot.
    $endgroup$
    – Aryan Pandey
    Mar 20 at 12:04












    $begingroup$
    You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
    $endgroup$
    – Cameron Buie
    Mar 20 at 12:36




    $begingroup$
    You're very welcome! I just realized there was something grievously wrong with my answer. It should be fixed, now.
    $endgroup$
    – Cameron Buie
    Mar 20 at 12:36











    1












    $begingroup$

    Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      How you got idea , to use this expression?
      $endgroup$
      – Aryan Pandey
      Mar 20 at 10:39















    1












    $begingroup$

    Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      How you got idea , to use this expression?
      $endgroup$
      – Aryan Pandey
      Mar 20 at 10:39













    1












    1








    1





    $begingroup$

    Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.






    share|cite|improve this answer









    $endgroup$



    Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 20 at 10:29









    HAMIDINE SOUMAREHAMIDINE SOUMARE

    1




    1











    • $begingroup$
      How you got idea , to use this expression?
      $endgroup$
      – Aryan Pandey
      Mar 20 at 10:39
















    • $begingroup$
      How you got idea , to use this expression?
      $endgroup$
      – Aryan Pandey
      Mar 20 at 10:39















    $begingroup$
    How you got idea , to use this expression?
    $endgroup$
    – Aryan Pandey
    Mar 20 at 10:39




    $begingroup$
    How you got idea , to use this expression?
    $endgroup$
    – Aryan Pandey
    Mar 20 at 10:39











    0












    $begingroup$

    Hint



    Express $$i=e^ipiover2\-i=e^i3piover2$$then $$x_1=r_1e^itheta_1 , x_1^2=i\x_2=r_2e^itheta_2 , x^2_2=-i$$what are the possible values of $x_1+x_2$?






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Hint



      Express $$i=e^ipiover2\-i=e^i3piover2$$then $$x_1=r_1e^itheta_1 , x_1^2=i\x_2=r_2e^itheta_2 , x^2_2=-i$$what are the possible values of $x_1+x_2$?






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Hint



        Express $$i=e^ipiover2\-i=e^i3piover2$$then $$x_1=r_1e^itheta_1 , x_1^2=i\x_2=r_2e^itheta_2 , x^2_2=-i$$what are the possible values of $x_1+x_2$?






        share|cite|improve this answer









        $endgroup$



        Hint



        Express $$i=e^ipiover2\-i=e^i3piover2$$then $$x_1=r_1e^itheta_1 , x_1^2=i\x_2=r_2e^itheta_2 , x^2_2=-i$$what are the possible values of $x_1+x_2$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 20 at 10:29









        Mostafa AyazMostafa Ayaz

        18.2k31040




        18.2k31040





















            0












            $begingroup$

            Hint:



            If $(a+bi)^2=alpha+beta i$, then:
            $$a=pmsqrtfracalphapmsqrtalpha^2+beta^22, b=pmsqrtfrac-alphapmsqrtalpha^2+beta^22$$



            Can you use this to find $sqrt i$ and $sqrt-i$ in the form $a+bi$?



            If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Got it , thanks
              $endgroup$
              – Aryan Pandey
              Mar 20 at 11:00















            0












            $begingroup$

            Hint:



            If $(a+bi)^2=alpha+beta i$, then:
            $$a=pmsqrtfracalphapmsqrtalpha^2+beta^22, b=pmsqrtfrac-alphapmsqrtalpha^2+beta^22$$



            Can you use this to find $sqrt i$ and $sqrt-i$ in the form $a+bi$?



            If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Got it , thanks
              $endgroup$
              – Aryan Pandey
              Mar 20 at 11:00













            0












            0








            0





            $begingroup$

            Hint:



            If $(a+bi)^2=alpha+beta i$, then:
            $$a=pmsqrtfracalphapmsqrtalpha^2+beta^22, b=pmsqrtfrac-alphapmsqrtalpha^2+beta^22$$



            Can you use this to find $sqrt i$ and $sqrt-i$ in the form $a+bi$?



            If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!






            share|cite|improve this answer











            $endgroup$



            Hint:



            If $(a+bi)^2=alpha+beta i$, then:
            $$a=pmsqrtfracalphapmsqrtalpha^2+beta^22, b=pmsqrtfrac-alphapmsqrtalpha^2+beta^22$$



            Can you use this to find $sqrt i$ and $sqrt-i$ in the form $a+bi$?



            If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 20 at 10:52

























            answered Mar 20 at 10:47









            Rhys HughesRhys Hughes

            7,0701630




            7,0701630











            • $begingroup$
              Got it , thanks
              $endgroup$
              – Aryan Pandey
              Mar 20 at 11:00
















            • $begingroup$
              Got it , thanks
              $endgroup$
              – Aryan Pandey
              Mar 20 at 11:00















            $begingroup$
            Got it , thanks
            $endgroup$
            – Aryan Pandey
            Mar 20 at 11:00




            $begingroup$
            Got it , thanks
            $endgroup$
            – Aryan Pandey
            Mar 20 at 11:00

















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