What is the amplitude and argument of the given complex number below?
$ z= 1-\cos(9\pi/10)-i \cdot\sin(9\pi/10)$
I have tried this a few times but the answer won't match the answer to this problem which is $-\pi/20$ for the argument.
This is my solution:
On
Draw a picture in the complex plane of a triangle with one vertex at $0$, one at $1$ and one at $\cos{{9\pi\over10}}+i\sin{{9\pi\over10}}$ Do you see it now?
On
Let $\theta=9\pi/20$ so $$z=1-\cos 2\theta-i\sin 2\theta =2\sin\theta (\sin\theta -i\cos\theta)=-2i\sin\theta\exp i\theta,$$which has modulus $|2\sin\theta|$ and argument $\theta-\frac{\pi}{2}\operatorname{sign}\sin\theta$. For the problem at hand, $\operatorname{sign}\sin\theta=1$.
On
If the argument is $\theta$, and the point is in the fourth quadrant so $\theta = \arctan \frac{-\sin \frac {9}{10}\pi}{1 - \cos \frac 9{10}{\pi}}$ which if one is too lazy to do trig you can just punch into a calculator to get is $-.05 \pi$.
Or I could remember my trig identities.
$1-\cos \frac 9{10}\pi = 1-\cos 2\frac 9{20}\pi = 1 - \cos^2 \frac 9{20}\pi + \sin^2 \frac 9{20}\pi = 2\sin^2 \frac 9{20}\pi$.
And $\sin \frac 9{10}\pi = \sin 2\frac 9{20}\pi = 2\sin \frac 9{20}\pi\cos \frac 9{20}\pi$.
So $\theta = \arctan \frac{-2\sin \frac 9{20}\pi\cos \frac 9{20}\pi}{ 2\sin^2 \frac 9{20}\pi} = $
$\arctan \frac{-\cos\frac 9{20}\pi}{\sin \frac 9{20}\pi}=$
$\arctan \cot -\frac 9{20}\pi=$
$-(\pi - \frac 9{20}\pi)=-\frac 1{20}\pi$
Or I could draw the picture below.
Anyway the argument is $-\frac 1{20}\pi$.
$1-e^{i\theta}=e^{i\frac{\theta}{2}} e^{-i\frac{\theta}{2}}-e^{i\frac{\theta}{2}} e^{i\frac{\theta}{2}}$ $= -e^{i\frac{\theta}{2}}(e^{i\frac{\theta}{2}}- e^{-i\frac{\theta}{2}})=-2i\sin{\frac{\theta}{2}} e^{i\frac{\theta}{2}}.$
$z=1-e^{\frac{9i\pi}{10}}=-2i\sin{\frac{9\pi}{20}}e^{\frac{9i\pi}{20}}.$
Hence the magnitude is $ 2\sin{\frac{9\pi}{20}}$ and the argument $-\frac{\pi}{2}+\frac{9\pi}{20}=\frac{-\pi}{20}$