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=√((-3)^2+4^2)\end{cases}`

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

So you have


`\begin{cases}\cos x=-\frac35\\[1ex]\sin x=\frac45\end{cases}`

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


`(\sin x)/(\cos x)=\tan x=(\frac45)/(-\frac35)=-\frac43`

Because the argument's cosine is negative and its sine is positive, you know that
`\frac\pi2<x<\pi`. This is important to know because it's only the case that
`y=\tan x\implies \arctan y=x` whenever
`-\frac\pi2<x<\frac\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=-\frac43\implies x=\pi+\arctan\left(-\frac43\right)\approx2.2143\text{ rad}\approx126.9^\circ`

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