Answer to Question #262541 in Complex Analysis for luka

Question

Given two complex numbers z1=1+i and z2= √(3)-1

a) Write z1/z2 in algebraic and polar forms.

b) Deduce the exact values of Cos 5pie/12 and Sin 5pie/12

c) What is the lowest positive value of integer n such that (z1/z2) is real

Answer 1

a)

$$\dfrac{z_1}{z_2}=\dfrac{1+i}{\sqrt{3}-i}=\dfrac{(1+i)(\sqrt{3}+i)}{3-1}$$

$$=\dfrac{\sqrt{3}+i+i\sqrt{3}-1}{4}=\dfrac{\sqrt{3}-1}{4}+\dfrac{\sqrt{3}+1}{4}i$$

$$z_1=1+i, |z|=\sqrt{1^2+1^2}=\sqrt{2}$$

$$\tan\theta_1=\dfrac{1}{1}=1$$

$$z_1=1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$$

$$z_2=\sqrt{3}-i, |z_2|=\sqrt{(\sqrt{3})^2+(-1)^2}=2$$

$$\tan\theta_2=\dfrac{-1}{\sqrt{3}}, \theta_2=-\pi/6$$

$$z_2=\sqrt{3}-i=2(\cos(-\pi/6)+i\sin(-\pi/6))$$

$$\dfrac{z_1}{z_2}=\dfrac{1+i}{\sqrt{3}-i}=\dfrac{\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))}{2(\cos(-\pi/6)+i\sin(-\pi/6))}$$

$$=\dfrac{\sqrt{2}}{2}(\cos(\pi/4+\pi/6)+i\sin(\pi/4+\pi/6))$$

$$=\dfrac{\sqrt{2}}{2}(\cos(5\pi/12)+i\sin(5\pi/12))$$

b)

$$\dfrac{\sqrt{3}-1}{4}=\dfrac{\sqrt{2}}{2}\cos(5\pi/12)$$

$$\cos(5\pi/12)=\dfrac{\sqrt{6}-\sqrt{2}}{4}$$

$$\dfrac{\sqrt{3}+1}{4}=\dfrac{\sqrt{2}}{2}\sin(5\pi/12)$$

$$\sin(5\pi/12)=\dfrac{\sqrt{6}+\sqrt{2}}{4}$$

c)

$$(\dfrac{z_1}{z_2})^n=(\dfrac{\sqrt{2}}{2})^n(\cos(\dfrac{5\pi n}{12})+i\sin(\dfrac{5\pi n}{12}))$$

If $(\dfrac{z_1}{z_2})^n$ is real, then $\sin(\dfrac{5\pi n}{2})=0.$

$$\dfrac{5\pi n}{12}=\pi m, m\in \Z$$

$$n=\dfrac{12m}{5}$$

The lowest positive value of integer $n$ is $n=12.$