Answer to Question #220737 in Complex Analysis for pappy

Question

Let z1 =-i/1+i,z2 = 1+i/1-i,and z3 = 1/10[2(i-1)1 +(i +√3)³+(1- i)(1-i)].Express z1z3/z2,z1z2/z3,and z1/z3z2 in both polar and standard forms.

Answer 1

$$z_1=-\dfrac{i}{1+i}, z_2=\dfrac{1+i}{1-i},$$

$$z_3=\dfrac{1}{10}[2(i-1)(1-i) +(i +√3)^3+(1- i)(1-i)]$$

$$z_1=-\dfrac{i}{1+i}=-\dfrac{i(1-i)}{2}=-\dfrac{1}{2}-\dfrac{1}{2}i$$

$$z_2=\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{2}=i$$

$$(i+\sqrt{3})^3=-i-3\sqrt{3}+9i+3\sqrt{3}=8i$$

$$(1-i)(1-i)=1-2i-1=-2i$$

$$2(i-1)(1-i)=4i$$

$$z_3=\dfrac{1}{10}[4i+8i-2i]=i$$

$$\dfrac{z_1z_3}{z_2}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)i}{i}=-\dfrac{1}{2}-\dfrac{1}{2}i$$

$$=\dfrac{\sqrt{2}}{2}(\cos(-\dfrac{3\pi}{4})+i\sin(-\dfrac{3\pi}{4}))$$

$$\dfrac{z_1z_2}{z_3}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)i}{i}=-\dfrac{1}{2}-\dfrac{1}{2}i$$

$$=\dfrac{\sqrt{2}}{2}(\cos(-\dfrac{3\pi}{4})+i\sin(-\dfrac{3\pi}{4}))$$

$$\dfrac{z_1}{z_2z_3}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)}{(i)i}=\dfrac{1}{2}+\dfrac{1}{2}i$$

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