Find the principal argument of $1+\cos\frac{11\pi}{9} + i\sin\frac{11\pi}{9}$

Question

If I write 1 as $\cos 0 + i\sin 0$

The expression becomes $$2\cos\frac{11\pi}{18} \left(\cos\frac{11\pi}{18}+i\sin\frac{11\pi}{18}\right)$$ The argument can be separated. So the argument for first part will be $-\pi$, or rather should be, because I am confused about the coming steps. What should I do next?

Answer 1

Evaluate first,

$$\frac{\sin\frac{11\pi}{9}}{1+\cos\frac{11\pi}{9}} =\frac{2\sin\frac{11\pi}{18}\cos\frac{11\pi}{18}}{2\cos^2\frac{11\pi}{18}} =\frac{\sin\frac{11\pi}{18}}{\cos\frac{11\pi}{18}}=\tan\frac{11\pi}{18}$$

So, since $\sin\frac{11\pi}{9}$ is negative, i.e. the 4th quadrant, the principal argument is $\frac{11\pi}{18}-\pi = -\frac{7\pi}{18} $

Or, continue with your expression to get the same result,

$$2\cos\frac{11\pi}{18} \left(\cos\frac{11\pi}{18}+i\sin\frac{11\pi}{18}\right) =2\cos\frac{7\pi}{18}e^{-i\pi}e^{i\frac{11\pi}{18}} =2\cos\frac{7\pi}{18}e^{-i\frac{7\pi}{18}} $$

Thus, $ -\frac{7\pi}{18}$.

Edit: Maybe it is implicit to some, but as Dr Zafar Ahmed DSc alluded below, the principal argument is defined within $(-\pi,\pi]$

Answer 2

Let $ O $ be the origin, $ C $ be 1, $ P = 1 + \cos\left(\frac{11\pi}{9}\right) + i \sin \left(\frac{11\pi}{9} \right) $, and $ R $ be 2, all in the complex plane. Then, if you draw a picture of this, you see that $ O, P, R $ are all points on the circle of radius 1 centered at $ C $. $ \angle RCP $ is $ -\frac{7\pi}{9} $. By a basic result in geometry on inscribed angles in a circle, $ \angle ROP = \frac{1}{2} \angle RCP = -\frac{7\pi}{18} $. Here I am using the convention of negative numbers to denote clockwise angles. I edited this answer because I did not notice the clockwise angles when I looked at this before.

Answer 3

The principal value of Arg lies in $(-\pi,\pi]$ and $e^{2ni\pi}=1, e^{i\pi}=-1.$

Given that $$Z=2 \cos 110^0 ~e^{11i \pi/18}= |2\cos 110^0|~ (1) e^{11i\pi/18} \implies Arg Z= \pi+11\pi/18$$ in order to bring this value in $(-\pi, \pi]$ we add $2n\pi$ to it where $n=\pm 1, \pm 2, \pm 3,...$, Finally, the principal value of $Arg Z$ is $$\pi+ \frac{11 \pi}{18}-2\pi=- \frac{7 \pi}{18}$$

Answer 4

$$\cos\dfrac{11\pi}{18}=\cos\left(\pi-\dfrac{7\pi}{18}\right)=-\cos\dfrac{7\pi}{18}<0$$

$$\sin\dfrac{11\pi}{18}=\sin\left(\pi-\dfrac{7\pi}{18}\right)=\sin\dfrac{7\pi}{18}>0$$

Using atan2

the principal argument will be $$\arctan\left(\dfrac{-2\cos\dfrac{7\pi}{18}\sin\dfrac{7\pi}{18}}{2^2\cos\dfrac{7\pi}{18}}\right)=\arctan\left(\tan\left(-\dfrac{7\pi}{18}\right)\right)=-\dfrac{7\pi}{18}$$