# eulers

• Aug 7th 2009, 09:53 PM
acosta0809
eulers
The question is

http://www.webassign.net/www22/latex...d58f9d7ef1.gif

im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

thank you
• Aug 7th 2009, 10:12 PM
red_dog
$\displaystyle i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$

$\displaystyle z=\sqrt{i}=\cos\frac{\frac{\pi}{2}+2k\pi}{2}+i\sin \frac{\frac{\pi}{2}+2k\pi}{2}, \ k=0,1$

$\displaystyle k=0\Rightarrow z=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{\sqrt {2}}{2}+\frac{\sqrt{2}}{2}i$

$\displaystyle k=1\Rightarrow z=\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$
• Aug 7th 2009, 10:26 PM
Failure
Quote:

Originally Posted by acosta0809
The question is

http://www.webassign.net/www22/latex...d58f9d7ef1.gif

im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

thank you

You don't get a unique answer to this:

a) $\displaystyle \sqrt{\mathrm{i}}=\left(\mathrm{e}^{\mathrm{i}\lef t(\frac{\pi}{2}+n\cdot 2\pi\right)}\right)^{1/2}=$(Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{4}+n\cdot\p i\right)}$ where $\displaystyle n\in \mathbb{Z}$
So, basically you have the two answers $\displaystyle \sqrt{\mathrm{i}}=$(Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$ and $\displaystyle \sqrt{\mathrm{i}}=$ (Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}$
Therefore, wirting an equality sign $\displaystyle \sqrt{\mathrm{i}}=\ldots$ seems rather problematic. Rather, we would want to say that the equation $\displaystyle z^2=\mathrm{i}$ has two solutions, $\displaystyle z_1=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$ and $\displaystyle z_2=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}$.

b) Just use $\displaystyle r\cdot\mathrm{e}^{\mathrm{i\varphi}}=r\cdot\cos(\v arphi)+\mathrm{i}r\cdot \sin(\varphi)$ for the above two solutions $\displaystyle z_1, z_2$.
• Aug 7th 2009, 11:46 PM
alunw
Complex square root without trig
Quote:

Originally Posted by Failure
You don't get a unique answer to this:

a) $\displaystyle \sqrt{\mathrm{i}}=\left(\mathrm{e}^{\mathrm{i}\lef t(\frac{\pi}{2}+n\cdot 2\pi\right)}\right)^{1/2}=$(Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{4}+n\cdot\p i\right)}$ where $\displaystyle n\in \mathbb{Z}$
So, basically you have the two answers $\displaystyle \sqrt{\mathrm{i}}=$(Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$ and $\displaystyle \sqrt{\mathrm{i}}=$ (Thinking) $\displaystyle \mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}$
Therefore, writing an equality sign $\displaystyle \sqrt{\mathrm{i}}=\ldots$ seems rather problematic. Rather, we would want to say that the equation $\displaystyle z^2=\mathrm{i}$ has two solutions, $\displaystyle z_1=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$ and $\displaystyle z_2=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}$.

b) Just use $\displaystyle r\cdot\mathrm{e}^{\mathrm{i\varphi}}=r\cdot\cos(\v arphi)+\mathrm{i}r\cdot \sin(\varphi)$ for the above two solutions $\displaystyle z_1, z_2$.

a)Almost every number has two distinct square roots. But $\displaystyle \surd 2$ conventionally denotes the positive solution to $\displaystyle z^2=2$. So the usual convention for square roots of complex numbers is to say that $\displaystyle \surd x$ is the square root with a positive real part. If x is real and negative then $\displaystyle \surd x$ is the root with positive imaginary part. This means there is a branch cut along the negative real axis.

b) In fact you can find find square roots without resorting to trigonometry. The following pseudo-code, computes the square root (x,y) of the complex number (u,v)
Code:

if (u > 0) then
x = sqrt((u+sqrt(u*u+v*v))/2)
y = v/(2*x)
else if (u < 0) then
y = sign(v)*sqrt((-u+sqrt(u*u+v*v))/2)
x = v/(2*y)
else
x = sqrt(abs(v)/2)
if (v > 0)
y = x
else
y = -x

The point of testing the sign of u is to minimise round-off error.