Answer to Question #117388 in Complex Analysis for Amoah Henry

Question

9. Find the moduli and principal arguments of w, z, wz and w/z, given that (a) w = 10i, z = 1 + i
3 + 2i, z = 1 − i, (c) w = , . 1 1 + i √

Answer 1

a)

$$w=10i, |w|=10,$$ $$Arg(w)={\pi \over 2}$$

$$z=1+\sqrt{3}i, |z|=\sqrt{(1)^2+(\sqrt{3})^2}=2,$$ $$Arg(z)={\pi \over 3}$$

$$|wz|=|w||z|=10(2)=20,$$ $$Arg(wz)=Arg(w)+Arg(z)={\pi \over 2}+{\pi \over 3}={5\pi \over6}$$

$$|{w \over z}|={|w|\over |z|}={10 \over 2}=5,$$ $$Arg({w \over z})=Arg(w)-Arg(z)={\pi \over 2}-{\pi \over 3}={\pi \over6}$$

b)

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

$$|w|=\sqrt{({1 \over 2})^2+({\sqrt{3} \over 2})^2}=1, Arg(w)=-{\pi\over 3}$$

$$z=1-i, |z|=\sqrt{(1)^2+(-1)^2}=\sqrt{2},$$ $$Arg(z)=-{\pi \over 4}$$

$$|wz|=|w||z|=1(\sqrt{2})=\sqrt{2},$$ $$Arg(wz)=Arg(w)+Arg(z)=-{\pi \over 3}-{\pi \over 4}=-{7\pi \over12}$$

$$|{w \over z}|={|w|\over |z|}={1 \over \sqrt{2}}={\sqrt{2} \over 2},$$

$$Arg({w \over z})=Arg(w)-Arg(z)=-{\pi \over 3}-(-{\pi \over 4})=-{\pi \over12}$$