argument of the complex number $1-\cos x-i\sin x$

Question

How can I find the argument of $1-\cos(x)-i\sin(x)$? Can I use the exponential form of a complex number? If so, how do I continue to find the the answer?

Answer 1

You can use the exponential form of a complex number. In those terms the argument is $\phi$: $$ re^{i\phi}=1-e^{ix}. $$ Then you can just say $re^{i\phi}=r\cos\phi+ir\sin\phi$ and use trigonometry. You have to solve $r\sin\phi=-\sin x$ and $r\cos\phi=1-\cos x$, so first solve for $r$ by eliminating $\phi$, then solve for $\phi$.

Answer 2

\begin{equation*} \begin{split} 1-\cos x -i\sin x & = 2\sin^2\frac{x}{2}-2i\sin\frac{x}{2}\cos\frac{x}{2} \\ & = 2\sin\frac{x}{2}\left[\cos\left(\frac{x}{2}-\frac{\pi}{2}\right) + i\sin \left(\frac{x}{2}-\frac{\pi}{2}\right)\right] \end{split} \end{equation*}

Answer 3

Well, you can start by noting that, if $w:=1-e^{ix},$ then whatever value of $x$ you choose, $w$ will have a non-negative real part. (Why?) Moreover, the real part of $w$ will be zero iff $w=0,$ which happens iff $x$ is an integer multiple of $2\pi.$ Otherwise, we can simply consider $$\arctan\left(-\frac{\sin x}{1-\cos x}\right)$$ in light of the half-angle formulae, the fact that the arctangent function is odd, and the fact that the tangent function's fundamental period is $\pi.$

Answer 4

As $$1-\cos x-i\sin x=2\sin^2\frac x2-i2\sin\frac x2\cos\frac x2$$

the argument will be undefined if $\sin\dfrac x2=0\iff \dfrac x2=m\pi\iff x=2m\pi$ where $m$ is any integer

Else let $1-\cos x=r\cos A,-\sin x=r\sin A$ where $-\pi<A\le2\pi,r\ge0$

arg$(1-\cos x-i\sin x)=\dfrac{r\sin A}{r\cos A}=\dfrac{-2\sin\dfrac x2\cos\dfrac x2}{2\sin^2\frac x2}=-\cot\dfrac x2=\tan\left(\dfrac\pi2+\dfrac x2\right)$

$\implies A=n\pi+\dfrac\pi2+\dfrac x2$ where $n$ is any integer

As $1-\cos x\ge0,$

Case $\#1:$ if $-\sin x\ge0\iff\sin x\le0$

the argument will lie in the first Quadrant, so we need to find integer $n$ such that $0\le A\le\dfrac{\pi}2$

Case $\#2:$ if $-\sin x<0\iff\sin x>0$

the argument will lie in the four Quadrant, so we need to find integer $n$ such that $-\dfrac{\pi}2\le A<0$