Question

(a) $1+i\tan \alpha (-\pi <\alpha <\pi ,\alpha \ne \pm \frac{\pi }{2})$
(b) Find the modulus and argument of the complex number $z_1=z^2-z$ if $z=\cos \varphi +i\sin \varphi$.

Answer 1

Let $r\cos \theta =1$, $r\sin \theta =\tan \alpha$
Then $r=\sqrt{1+{\tan }^2\alpha }=\sec \alpha =\frac{1}{\left|\cos \theta \right|}$
Then $\cos \theta =\left|\cos \theta \right|$ .....(1)
$\sin \theta =\tan \alpha \left|\cos \alpha \right|$ .....(2)
b) Substituting $z=\cos \varphi +i\sin \varphi$, we get
$z_1={\left(\cos \varphi +i\sin \varphi \right)}^2-\left(\cos \varphi +i\sin \varphi \right)$
$=\cos 2\varphi +i\sin 2\varphi -\cos \varphi -i\sin \varphi$
$=\left(\cos 2\varphi -\cos \varphi \right)+i\left(\sin 2\varphi -\sin \varphi \right)$
$2\sin \frac{3\varphi }{2}\sin (-\frac{\varphi }{2})+2i\cos \frac{3\varphi }{2}\sin \frac{\varphi }{2}$
$2\sin \frac{\varphi }{2}[-\sin \left(\frac{3\varphi }{2}\right)+i\cos \frac{3\varphi }{2}]$
$2\sin \frac{\varphi }{2}[(\frac{\pi }{2}+\frac{3\varphi }{2})+i\sin (\frac{\pi }{2}+\frac{3\varphi }{2})]$
$\left|z\right|=2\left|\sin \frac{\varphi }{2}\right|$
In view of the definition of modulus, we consider three cases:
Case I. Let $\sin \frac{\varphi }{2}=0$, that is, $\varphi =2n\pi$, $n$ any integer. Than $\left|z_1\right|=0$ which implies that $z_1=0$. Thus, for $\varphi =2n\pi$ ($n$, any integer), arg $z_1$ is undefined.
Case II. Let $\sin \frac{\varphi }{2}>0$ which occurs when
$0<\frac{\varphi }{2}<\pi$ i.e. I, II Quadrant.
or $2n\pi <\frac{1}{2}\varphi <(2n+1)\pi$, that is, when
$4n\pi <\varphi <(4n+2)\pi$, $n$ any integer. .....(1)
Then $\left|z_1\right|=2\sin \frac{\pi }{2}$ and the trigonometrical form of $z_1$ is as
$z_1=2\sin \frac{\varphi }{2}\left[\cos \left(\frac{\pi +3\varphi }{2}\right)+i\sin \left(\frac{\pi +3\varphi }{2}\right)\right]$
Consequently, if $\varphi$ satisfies the condition $\left(1\right)$,
then $\arg z_1=\frac{\pi +3\varphi }{2}.$
Case III. Let $\sin \frac{\varphi }{2}<0$, that is,
$\pi <\frac{\varphi }{2}<2\pi$ i.e. III, IV Quadrant.
or $(2n+1)\pi <\frac{1}{2}\varphi <(2n+2)\pi$
or $(4n+2)\pi <\varphi <(4n+4)\pi$, .....(2)
$n$ any integer.
Then $\left|z_1\right|=-2\sin \frac{\varphi }{2}$ and the trigonometrical form of $z_1$ is $-2\sin \frac{\varphi }{2}\left[\sin \frac{3\varphi }{2}-i\cos \frac{3\varphi }{2}\right]$ by $\left(A\right)$
$z_1=-2\sin \frac{\varphi }{2}\left[\cos \frac{3\pi +3\varphi }{2}+i\sin \frac{3\pi +3\varphi }{2}\right]$
Hence, when $\varphi$ satisfies $\left(2\right)$, we have
$\arg z_1=\left(\frac{3\pi +3\varphi }{2}\right).$