[tex]-3+4i[/tex] is a complex number that satisfies
[tex]\begin{cases}r\cos x=-3\\[1ex]r\sin x=4\\[1ex]r=\sqrt{(-3)^2+4^2}\end{cases}[/tex]
The last equation immediately tells you that [tex]r=5[/tex].
So you have
[tex]\begin{cases}\cos x=-\dfrac35\\[1ex]\sin x=\dfrac45\end{cases}[/tex]
Dividing the second equation by the first, you end up with
[tex]\dfrac{\sin x}{\cos x}=\tan x=\dfrac{\dfrac45}{-\dfrac35}=-\dfrac43[/tex]
Because the argument's cosine is negative and its sine is positive, you know that [tex]\dfrac\pi2<x<\pi[/tex]. This is important to know because it's only the case that [tex]y=\tan x\implies \arctan y=x[/tex] whenever [tex]-\dfrac\pi2<x<\dfrac\pi2[/tex]. 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 [tex]\pi[/tex], we have
[tex]\tan(x-\pi)=y\implies x=\pi+\arctan y[/tex]
All this to say
[tex]\tan x=-\dfrac43\implies x=\pi+\arctan\left(-\dfrac43\right)\approx2.2143\text{ rad}\approx126.9^\circ[/tex]
So, [tex]-3+4i=5(\cos126.9^\circ+i\sin126.9^\circ)[/tex].
