$\arg(z_1 + z_2)$ if $z_1 = 2cis\left(\frac{\pi}{12}\right)$ and $z_2 = 2i$

Question

$\arg(z_1 + z_2)$ if $z_1 = 2cis\left(\frac{\pi}{12}\right)$ and $z_2 = 2i$

I have attempted to put both in the same form, but am stuck. Any help would be greatly appreciated!

Thank you in advance.

Answer 1

If $z_1=2~ cis(\pi/12):$

$$z=z_1+z2=2\cos(\pi/12)+2i\sin(\pi/12)+2i$$ Let $\frac{\pi}{12}=t$ $$z=2[\cos t+i (1+\sin t)]$$ $$\implies z=2[\{\cos^2(t/2)-\sin^2(t/2)\}+i\{(\sin(t/2)+\cos(t/2)\}^2]$$

$$\implies z=2[\cos (t/2) + \sin (t/2)]~[\{\cos(t/2)-\sin(t/2)\}+i\{(\sin(t/2)+\cos(t/2)\}]$$ The firsyt sq. Barket is positive as $t/2$ is acute angle. The argument of the complex number in the next Sq. brackets will be thae same as that of $z$ $$ Arg(z)=\tan^{-1} \frac{1+\tan(t/2)}{1-\tan(t/2)}= \pi/2+t/2=\pi/4+\pi/24 =\frac{7\pi}{24}.$$

Answer 2

$$z=2\cos(\pi/12)+2i \implies Arg (z)=\tan^{-1} \sec(\pi/12)$$ $$Arg(z)=\tan^{-1}(\sqrt{6}-\sqrt{2}) \approx \tan^{-1} (1.03528) \approx \pi/4$$ Here we use $$\cos(\pi/12)=\frac{1+\sqrt{3}}{2\sqrt{2}}$$

Answer 3

Note, $z_1 = 2cis\frac{\pi}{12}=2e^{i \frac{\pi}{12}}$ and $z_2 = 2i=2e^{i\frac\pi2}$. Let $\arg(z_1 + z_2)=\theta$. Then,

$$z_1+z_2 = 2 (e^{i\frac\pi2}+e^{i \frac{\pi}{12}}) =2e^{i \theta}[e^{i(\frac\pi2-\theta)}+e^{-i (\theta-\frac{\pi}{12})}]$$

For $e^{i(\frac\pi2-\theta)}+e^{-i (\theta-\frac{\pi}{12})}$ to be real, set

$$\frac\pi2-\theta = \theta-\frac{\pi}{12}$$

which yields $\arg(z_1 + z_2)=\theta=\frac{7\pi}{24}$ and

$$z_1+z_2 =4\cos\frac{5\pi}{24}e^{i \frac{7\pi}{24}}$$