In schaumm's series book for complex variables, I found this question

Why this fallany happens

-1=\sqrt(-1)*\sqrt(-1)=\sqrt(-1*-1)=\sqrt(1)=1

Could someone help me find the error?

Thank you so much.

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- Jun 13th 2013, 12:50 AMerich22Complex Variables cocerning i
In schaumm's series book for complex variables, I found this question

Why this fallany happens

-1=\sqrt(-1)*\sqrt(-1)=\sqrt(-1*-1)=\sqrt(1)=1

Could someone help me find the error?

Thank you so much. - Jun 13th 2013, 02:50 AMPlatoRe: Complex Variables cocerning i
You are applying multiplication rules that do not apply to complex numbers.

If $\displaystyle z=r\exp(i\theta)~~w=s\exp(i\phi)$ then $\displaystyle z\cdot w=rs\exp(i(\theta+\phi))$.

Now note that $\displaystyle i=\exp\left(\frac{i\pi}{2}\right)$ so $\displaystyle i\cdot i=?$ - Jun 13th 2013, 04:19 AMHallsofIvyRe: Complex Variables cocerning i
Specifically, $\displaystyle \sqrt{a}\sqrt{b}= \sqrt{ab}$ is not true for complex numbers.

- Jun 13th 2013, 11:23 AMerich22Re: Complex Variables cocerning i
I started to think, Plato....

I think it's maybe the \sqrt(1) is not just 1 depends on which branch are you on, isn't it?

What do you think...? - Jun 13th 2013, 11:27 AMerich22Re: Complex Variables cocerning i
Ok, HallsofIvy, if it is not true, so we can not calculate \sqrt(-4)=\sqrt(4*-1)=\sqrt(4)*\sqrt(-1)=2i, can't we?

- Jun 13th 2013, 11:40 AMPlatoRe: Complex Variables cocerning i
No, $\displaystyle \sqrt{1}$ is a real number and as such it has only one value $\displaystyle \sqrt{1}=1$.

The complex numbers are enlargement of the real numbers.

The number $\displaystyle i$ is introduced as the root of the equation $\displaystyle x^2+1=0$.

You can see that then $\displaystyle i^2$ must $\displaystyle =-1$.

So it is easy to see why in history it was tempting to say $\displaystyle i=\sqrt{-1}$. - Jun 13th 2013, 11:53 AMPlatoRe: Complex Variables cocerning i
- Jun 13th 2013, 01:17 PMerich22Re: Complex Variables cocerning i
I'm rather confused, Plato.

In the complex system, is \sqrt(1) means we want to find complex numbers z such that z^2=1?

That means in complex system, the \sqrt(1) = +1 and -1, doesn't it?

It differs from the real system that defines \sqrt(1) as a positive number x such that x^2=1. Thus the solution is just x=1, without x=-1?

CMIW - Jun 13th 2013, 01:24 PMerich22Re: Complex Variables cocerning i
Plato, concerning \sqrt{-4}=\sqrt{4\cdot-1}=\sqrt{4}*\sqrt{-1}=2i.

But is it true when we want to find for example the solution of z^2-i*z+2=0, then we could use

z_12=(i \pm \sqrt((-i)^2-4*1*2))/2=(i \pm \sqrt(-9))/2=(i \pm \sqrt(9*(-1)))/2=(i \pm \sqrt(9)*\sqrt(-1))/2=2i or -i? - Jun 13th 2013, 01:49 PMPlatoRe: Complex Variables cocerning i
Absolutely NOT! Nowhere, is that true.

$\displaystyle \sqrt{1}=1,~\sqrt{1}\ne -1,~\&~-\sqrt{1}=-1$

The fact is: the use of a radical is for real numbers.

Now it is true that for historical reasons we*wave our hands*and allow $\displaystyle \sqrt{-9}=3i$.

BUT**the rules for radicals apply only to real numbers**and do not carry over to complex.

Example: $\displaystyle \sqrt{-3}\cdot\sqrt{-3}=\sqrt{3}i\cdot\sqrt{3}i=3i^2=-3$,*(hand waving)*.

Because $\displaystyle \sqrt{-3}$ is not real, we do not apply rules for radicals. - Jun 13th 2013, 09:53 PMerich22Re: Complex Variables cocerning i
Thank you so much, Plato. Now I understand.

One more question, Plato. But what about the function f(z)=\sqrt(z)? Does it exist? - Jun 14th 2013, 02:44 AMProve ItRe: Complex Variables cocerning i
Yes, it exists. The idea with complex functions is to write them in terms of their real and imaginary parts, each of which will be real functions.

Say $\displaystyle \displaystyle \begin{align*} z = x + i\,y = r\,e^{i \left( \theta + 2\pi n \right) } \textrm{ where } r = \sqrt{x^2 + y^2}, \, \theta = \textrm{Arg}\,{(z)} \textrm{ and } n \in \mathbf{Z} \end{align*}$, then

$\displaystyle \displaystyle \begin{align*} f(z) &= \sqrt{z} \\ &= \left[ r\, e^{i \left( \theta + 2\pi n \right) } \right] ^{\frac{1}{2}} \\ &= r^{\frac{1}{2}} \, e^{ i \left( \frac{ \theta }{2} + \pi n \right) } \\ &= \sqrt[4]{x^2 + y^2} \left\{ \cos{ \left[ \frac{\textrm{Arg}\,{(z)}}{2} + \pi n \right] } + i \sin{ \left[ \frac{\textrm{Arg}\,{(z)}}{2} + \pi n \right] } \right\} \\ &= \sqrt[4]{x^2 + y^2}\,\cos{\left[ \frac{\textrm{Arg}\,{(z)}}{2} + \pi n \right] } + i\,\sqrt[4]{x^2 + y^2}\,\sin{ \left[ \frac{\textrm{Arg}\,{(z)}}{2} + \pi n \right] } \end{align*}$ - Jun 14th 2013, 07:23 AMerich22Re: Complex Variables cocerning i
hmmm, but then 'Prove It', it means that f(1)=\sqrt(1)=1 or -1, doesn't it?