# Thread: help me derive this identity for complex numbers

1. ## help me derive this identity for complex numbers

Derive these identities.
1. $e^{iz} = cos z + isin z$
2. $arctan z = \frac{i}{2}log(\frac{i+z}{i-z})$ (show that the set {w:tan w=z} is the same as the set $\frac{i}{2}log(\frac{i+z}{i-z})$)
3. $\frac{d}{dz}arctan z = \frac{1}{1+z^2}$

2. Hello, splash!

It's not clear what definitions and theorems you are allowed.

1) Derive: $e^{iz} \,= \,\cos z + i\sin z$

I'll assume that we are allowed these two defintions:

. . $[1]\;\cos z \:=\:\frac{e^{iz} + e^{-iz}}{2}\qquad[2]\; \sin z \:= \:\frac{e^{iz} - e^{-iz}}{2i}$

Multiply [2] by $i:\;\;i\sin z\;=\;i\cdot\frac{e^{iz} - e^{-iz}}{2i}$

We have: . $i\sin z \;=\;\frac{e^{iz} - e^{-iz}}{2}$

Add [1]:. . $\cos z \;=\;\frac{e^{iz} + e^{-iz}}{2}$

Then: . $\cos z + i\sin z\;=\;\frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2} \;=\;\frac{e^{iz} + e^{-iz} + e^{iz} - e^{-iz}}{2} \;=\;\frac{2e^{iz}}{2}$

. . Therefore: . $\boxed{\cos z + i\sin z \;= \;e^{iz}}$

3. Originally Posted by splash
Derive these identities.
1. $e^{iz} = cos z + isin z$
One of the ways of deriving this is extending the definition of the exponent to a complex number.

It can be shown that for a real number, $r$ we have,
$e^r=1+r+\frac{r^2}{2!}+...$

So we define, an imaginary exponent,
$e^{zi}=1+(zi)+\frac{(zi)^2}{2!}+...$
(Withour fear of divergence, this fear is not only convergent but asbsolutely convergent. This fact will enable us to use Riemann's rearrangement theorem).

If we open parantheses we have,
$1+zi-\frac{z^2}{2!}-i\frac{z^3}{3!}+\frac{z^4}{4!}+i\frac{z^5}{5!}...$
Apply rearrangement and factorize,
$\left( 1-\frac{z^2}{2!}+\frac{z^4}{4!} \right)+i\left( z-\frac{z^3}{3!}+\frac{z^5}{5!}-...\right)$
You should realize these a power series expanstion for,
$\cos z+i\sin z$