Question

Write -3+4i in the form r(cosx+isinx)

Answer 1

$-3+4i$ is a complex number that satisfies

$\begin{cases}r\cos x=-3\\[1ex]r\sin x=4\\[1ex]r=\sqrt{(-3)^2+4^2}\end{cases}$

The last equation immediately tells you that $r=5$.

So you have

$\begin{cases}\cos x=-\dfrac35\\[1ex]\sin x=\dfrac45\end{cases}$

Dividing the second equation by the first, you end up with

$\dfrac{\sin x}{\cos x}=\tan x=\dfrac{\dfrac45}{-\dfrac35}=-\dfrac43$

Because the argument's cosine is negative and its sine is positive, you know that $\dfrac\pi2$. This is important to know because it's only the case that $y=\tan x\implies \arctan y=x$ whenever $-\dfrac\pi2$. The inverse doesn't exist otherwise.

However, you can restrict the domain of the tangent function so that an inverse can be defined. By shifting the argument of tangent by $\pi$, we have

$\tan(x-\pi)=y\implies x=\pi+\arctan y$

All this to say

$\tan x=-\dfrac43\implies x=\pi+\arctan\left(-\dfrac43\right)\approx2.2143\text{ rad}\approx126.9^\circ$

So, $-3+4i=5(\cos126.9^\circ+i\sin126.9^\circ)$.