The exponential form of $1-xe^{i \theta}$

Question

How can i find the exponential form of $1-xe^{i\theta}$, with $0 \le x\le 1$ and $0 < \theta < \frac{\pi}{2}$ . I tried to use the Moivre formula but the $x$ hinders my demo.

Answer 1

Hint: recall that $e^{i \theta}=\cos(\theta)+i\sin(\theta)$. Then, you need to recall that, given $z=a+ib$, $a,b \in \mathbb{R}$, its exponential form is $\rho e^{i \phi}$, where $\rho=\sqrt{a^2+b^2}$ and $\phi=\arctan \left ( {\frac{b}{a}} \right )$ (with the usual values in the case $a=0$).

Answer 2

Note

\begin{align} 1-xe^{i \theta}& = 1-x\cos \theta -i x\sin\theta =r e^{- i\alpha} \end{align}

where $r=\sqrt{1+x^2-2x\cos \theta } $ and $\alpha=\sin^{-1}(\frac{x}r\sin\theta$).