Solve for r
r=\left(6-3i\right)e^{-i\theta }
Solve for θ
\theta =-i\ln(\frac{6-3i}{r})+2\pi n_{1}
n_{1}\in \mathrm{Z}
r\neq 0
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\left(\cos(\theta )+i\sin(\theta )\right)r=6-3i
Combine all terms containing r.
\frac{\left(\cos(\theta )+i\sin(\theta )\right)r}{\cos(\theta )+i\sin(\theta )}=\frac{6-3i}{\cos(\theta )+i\sin(\theta )}
Divide both sides by \cos(\theta )+i\sin(\theta ).
r=\frac{6-3i}{\cos(\theta )+i\sin(\theta )}
Dividing by \cos(\theta )+i\sin(\theta ) undoes the multiplication by \cos(\theta )+i\sin(\theta ).
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