Complex number under Sqrt

Question

Is it possible that $\sqrt{a+ i \: b} $ could not be written as one of the below forms?

$ i \sqrt{-a-i\:b} $

or

$-i \sqrt{-a-i\:b}$

$a$ and $b$ are real constants.

It might be just me, but how are you saying that$$\sqrt{a+bi}=i\sqrt{-a-bi}$$And this question might be related to the common$$1=\sqrt{-1}\times\sqrt{-1}=i^2=-1$$fallacy. — Crescendo, Feb 26 '18 at 15:33

Dr. Sonnhard Graubner · Accepted Answer · 2018-02-26T15:29:23.673

2

set $$\sqrt{a+bi}=X+iY$$ and square it, then you will get $$a+bi=X^2-Y^2+2XYi$$ and it must be $$a=X^2-Y^2,b=2XY$$ now you can calculate $$X,Y$$

edited Feb 26 '18 at 15:29

answered Feb 26 '18 at 15:26

1

Unclear answer! – Wat Watson Feb 26 '18 at 15:27
why is my answer unclear? – Dr. Sonnhard Graubner Feb 26 '18 at 15:29
Because it doesn't provide a yes or no to the question – Wat Watson Feb 26 '18 at 15:32
Could you compute $$X,Y$$? – Dr. Sonnhard Graubner Feb 26 '18 at 15:44
The question is if $\sqrt{a+ib}$ equals $\pm i \sqrt{-a-ib}$, not if it can be written in the form $X+iY$. – Martin R Feb 26 '18 at 15:52
1

when you compute $X,Y$ then you will see it! – Dr. Sonnhard Graubner Feb 26 '18 at 15:53

Mehrdad Zandigohar · Answer 2 · 2018-02-26T16:20:05.463

1

$\displaystyle \sqrt{a+ i \: b}=\sqrt{-i^2(a+ i \: b)}=\sqrt{i^2} \cdot \sqrt{-(a+ i \: b)}=i\sqrt{-a-ib}$

Because: $\sqrt{i^2}=\sqrt{(\sqrt{-1})^2}=\sqrt{-1}=i$

Update: Thanks to Watson's comment as we can also find here we can't always separate the square roots but here we can.

If $|Arg(i^2)+Arg(a+ib)|≥π$, then $\sqrt{i^2 (a+ib)}=\sqrt {i^2} \sqrt {a+ib}$ does not hold,

edited Feb 26 '18 at 16:20

answered Feb 26 '18 at 15:38

It may not be possible to separate the sqrt to two parts – Wat Watson Feb 26 '18 at 15:42
Why not? $\sqrt{fg}=(fg)^{1/2}=(f)^{1/2} \cdot (g)^{1/2}=\sqrt{f} \cdot \sqrt{g}$ @WatWatson – Mehrdad Zandigohar Feb 26 '18 at 15:44
The are not real numbers – Wat Watson Feb 26 '18 at 15:46
You can say $(fg)^{n}=f^n \cdot g^{n}$ for anything, do you have an example for opposite of this statement? @WatWatson – Mehrdad Zandigohar Feb 26 '18 at 15:47
Check https://math.stackexchange.com/questions/1481316/sqrtab-sqrta-sqrtb-for-complex-number-a-and-b?rq=1 – Wat Watson Feb 26 '18 at 15:47
You left a very interesting comment, thank you for that. but it is still ok, because: If $|Arg(i^2)+Arg(a+ib)|≥π$, then $\sqrt{i^2 (a+ib)}=\sqrt {i^2} \sqrt {a+ib}$ does not hold, but we have: $|Arg(i^2)+Arg(a+ib)|=|Arg(a+ib)|\leqπ$, So it is ok.Isnt' it? @WatWatson – Mehrdad Zandigohar Feb 26 '18 at 16:01
I updated the answer @WatWatson – Mehrdad Zandigohar Feb 26 '18 at 16:20

2 Answers2