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

User Frayal
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5 votes

-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).
User Nucleartux
by
8.0k points
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