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Solutions of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$

Locating possible complex solutions to an equation involving a square rootFinding the square root of a complex number - why two solutions instead of four?square root of $frac2+sqrt34$$(n-1)^textth$ root of conjugateWhy don't $sqrta*b=sqrta*sqrtb$ holds true when one of $a$ & $b$ is imaginary number?Solve $x^4 + 64 = 0$Roots of complex quadratic polynomialWhy $Releft(e^sqrt-log xright)=Imleft(e^sqrt-log xright)$ has no real solutions?Are there multiple solutions to a fractional root/radical (over the complex numbers)?Question about this exponent problem.

0

$begingroup$

Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$

I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :

beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign

Why am I missing the solution $dfrac-2i5$ in my attempt ?

If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.

complex-numbers

share|cite|improve this question

edited Mar 16 at 21:53

ss1729

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

$endgroup$

add a comment | 

0

$begingroup$

Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$

I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :

beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign

Why am I missing the solution $dfrac-2i5$ in my attempt ?

If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.

complex-numbers

share|cite|improve this question

edited Mar 16 at 21:53

ss1729

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

$endgroup$

add a comment | 

$begingroup$

Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$

I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :

beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign

Why am I missing the solution $dfrac-2i5$ in my attempt ?

If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.

complex-numbers

share|cite|improve this question

edited Mar 16 at 21:53

ss1729

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

$endgroup$

share|cite|improve this question

edited Mar 16 at 21:53

ss1729

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

share|cite|improve this question

edited Mar 16 at 21:53

ss1729

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

asked Mar 16 at 21:27

ss1729ss1729

2,04411124

1 Answer
1

active

oldest

votes

2

$begingroup$

Remember that $841$ has two square roots.

share|cite|improve this answer

answered Mar 16 at 21:29

ArthurArthur

120k7119203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
  $endgroup$
  – ss1729
  Mar 16 at 21:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But here, we don't. Otherwise the expression on the right has exactly one value.
  $endgroup$
  – enedil
  Mar 16 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
  $endgroup$
  – ss1729
  Mar 16 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
  $endgroup$
  – Arthur
  Mar 16 at 22:07
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Remember that $841$ has two square roots.

share|cite|improve this answer

answered Mar 16 at 21:29

ArthurArthur

120k7119203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
  $endgroup$
  – ss1729
  Mar 16 at 21:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But here, we don't. Otherwise the expression on the right has exactly one value.
  $endgroup$
  – enedil
  Mar 16 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
  $endgroup$
  – ss1729
  Mar 16 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
  $endgroup$
  – Arthur
  Mar 16 at 22:07
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Remember that $841$ has two square roots.

share|cite|improve this answer

answered Mar 16 at 21:29

ArthurArthur

120k7119203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
  $endgroup$
  – ss1729
  Mar 16 at 21:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But here, we don't. Otherwise the expression on the right has exactly one value.
  $endgroup$
  – enedil
  Mar 16 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
  $endgroup$
  – ss1729
  Mar 16 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
  $endgroup$
  – Arthur
  Mar 16 at 22:07
  
  

  
  
  
  

add a comment | 

$begingroup$

Remember that $841$ has two square roots.

share|cite|improve this answer

answered Mar 16 at 21:29

ArthurArthur

120k7119203

$endgroup$

share|cite|improve this answer

answered Mar 16 at 21:29

ArthurArthur

120k7119203

answered Mar 16 at 21:29

ArthurArthur

120k7119203

answered Mar 16 at 21:29

ArthurArthur

120k7119203