# Complex Variables cocerning i

• Jun 13th 2013, 01:50 AM
erich22
Complex 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, 03:50 AM
Plato
Re: Complex Variables cocerning i
Quote:

Originally Posted by erich22
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?

You are applying multiplication rules that do not apply to complex numbers.

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

Now note that $i=\exp\left(\frac{i\pi}{2}\right)$ so $i\cdot i=?$
• Jun 13th 2013, 05:19 AM
HallsofIvy
Re: Complex Variables cocerning i
Specifically, $\sqrt{a}\sqrt{b}= \sqrt{ab}$ is not true for complex numbers.
• Jun 13th 2013, 12:23 PM
erich22
Re: 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, 12:27 PM
erich22
Re: 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, 12:40 PM
Plato
Re: Complex Variables cocerning i
Quote:

Originally Posted by erich22
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...?

No, $\sqrt{1}$ is a real number and as such it has only one value $\sqrt{1}=1$.

The complex numbers are enlargement of the real numbers.
The number $i$ is introduced as the root of the equation $x^2+1=0$.
You can see that then $i^2$ must $=-1$.
So it is easy to see why in history it was tempting to say $i=\sqrt{-1}$.
• Jun 13th 2013, 12:53 PM
Plato
Re: Complex Variables cocerning i
Quote:

Originally Posted by erich22
so we can not calculate $\sqrt{-4}=\sqrt{4\cdot-1}=\sqrt{4}*\sqrt{-1}=2i$, can't we?

That is true: you cannot do that.

However, $(2i)^2=4i^2=-4$. Therefore, in the enlarged system we can say $2i$ is the principal square root of $-4$.
We just don't use a radical symbol.
• Jun 13th 2013, 02:17 PM
erich22
Re: 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, 02:24 PM
erich22
Re: 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, 02:49 PM
Plato
Re: Complex Variables cocerning i
Quote:

Originally Posted by erich22
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?

Absolutely NOT! Nowhere, is that true.

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

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 $\sqrt{-9}=3i$.
BUT the rules for radicals apply only to real numbers and do not carry over to complex.

Example: $\sqrt{-3}\cdot\sqrt{-3}=\sqrt{3}i\cdot\sqrt{3}i=3i^2=-3$, (hand waving).
Because $\sqrt{-3}$ is not real, we do not apply rules for radicals.
• Jun 13th 2013, 10:53 PM
erich22
Re: 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, 03:44 AM
Prove It
Re: Complex Variables cocerning i
Quote:

Originally Posted by erich22
Thank you so much, Plato. Now I understand.

One more question, Plato. But what about the function f(z)=\sqrt(z)? Does it exist?

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 \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 \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, 08:23 AM
erich22
Re: Complex Variables cocerning i
hmmm, but then 'Prove It', it means that f(1)=\sqrt(1)=1 or -1, doesn't it?