Polar form for complex number with variable exponent This might be an easy question but I'm having trouble showing that
$\left(1+\frac{i\theta}{m}\right)^m$ has the angle $
m\arctan\left(\frac{\theta}{m}\right)$ in polar form on the complex plane.
Edit: I'm using this to show Euler's formula, so I can't use the formula in the proof.
 A: Hint:
The first step is to write $z=1+i\frac{\theta}{m}$ in polar form $re^{i\psi}$ (I use $\psi$ instead of $\theta$ here, because $\theta$ has already been used.)
Remember, if $z=re^{i\psi}$, then $r=\lvert z\rvert$. In this case,
$$
r=\lvert z\rvert=\sqrt{1+\frac{\theta^2}{m^2}}.
$$
Factoring this out, we have written
$$
z=r\left(\frac{1}{\sqrt{1+\frac{\theta^2}{m^2}}}+i\frac{\frac{\theta}{m}}{\sqrt{1+\frac{\theta^2}{m^2}}}\right).
$$
Now, whatever angle $\psi$ we choose, we want this to look like $r(\cos\psi+i\sin\psi)$; so, we must have
$$
\cos\psi=\frac{1}{\sqrt{1+\frac{\theta^2}{m^2}}}\qquad\sin\psi=\frac{\frac{\theta}{m}}{\sqrt{1+\frac{\theta^2}{m^2}}}.
$$
Can you see how to use these to come up with an expression for $\tan\psi$? From there, can you see how to find $\psi$?
Having done this, the last step of the result is to compute the angle for $z^m$; this isn't so bad if you rewrite it as $(re^{i\psi})^m$ and try it from there.
A: To show that the arg of a product is the sum of the args, consider the tangent of the args:

$$
(x+iy)(u+iv)=(xu-yv)+i(yu+xv)
$$
$$
\tan(\arg((x+iy)(u+iv)))=\frac{yu+xv}{xu-yv}\tag{1}
$$

$$
\tan(\arg(x+iy))=\frac yx
$$
$$
\tan(\arg(u+iv))=\frac vu
$$
$$
\begin{align}
\tan(\arg(x+iy)+\arg(u+iv))
&=\frac{\frac yx+\frac vu}{1-\frac yx\frac vu}\\
&=\frac{yu+xv}{xu-yv}\tag{2}
\end{align}
$$

Now $\arg(1+\frac{i\theta}{m})=\tan^{-1}\left(\frac{\theta}{m}\right)$, so inductively, we have $\arg\left((1+\frac{i\theta}{m})^m\right)=m\tan^{-1}\left(\frac{\theta}{m}\right)$
