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 $Re(z)=fracsqrt22$
$begingroup$
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
complex-numbers
$endgroup$
add a comment |
$begingroup$
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
complex-numbers
$endgroup$
1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Mar 20 at 10:26
add a comment |
$begingroup$
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
complex-numbers
$endgroup$
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
complex-numbers
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 $$b=pmsqrtfracsqrt10+12.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.
$endgroup$
$begingroup$
How you got idea , to use this expression?
$endgroup$
– Aryan Pandey
Mar 20 at 10:39
add a comment |
$begingroup$
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 $x_1+x_2$?
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
Got it , thanks
$endgroup$
– Aryan Pandey
Mar 20 at 11:00
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$b=pmsqrtfracsqrt10+12.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$b=pmsqrtfracsqrt10+12.$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$b=pmsqrtfracsqrt10+12.$$
$endgroup$
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 $$b=pmsqrtfracsqrt10+12.$$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.
$endgroup$
$begingroup$
How you got idea , to use this expression?
$endgroup$
– Aryan Pandey
Mar 20 at 10:39
add a comment |
$begingroup$
Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.
$endgroup$
$begingroup$
How you got idea , to use this expression?
$endgroup$
– Aryan Pandey
Mar 20 at 10:39
add a comment |
$begingroup$
Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.
$endgroup$
Hint: $i=frac12(1+i)^2$ and $-i=frac12(1-i)^2$.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $x_1+x_2$?
$endgroup$
add a comment |
$begingroup$
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 $x_1+x_2$?
$endgroup$
add a comment |
$begingroup$
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 $x_1+x_2$?
$endgroup$
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 $x_1+x_2$?
answered Mar 20 at 10:29
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
$begingroup$
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!
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$begingroup$
Got it , thanks
$endgroup$
– Aryan Pandey
Mar 20 at 11:00
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
Got it , thanks
$endgroup$
– Aryan Pandey
Mar 20 at 11:00
add a comment |
$begingroup$
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!
$endgroup$
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!
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
add a comment |
$begingroup$
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
add a comment |
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