Answer to Question #117386 in Complex Analysis for Amoah Henry

Question

. Find the modulus and the principal argument of each of the given complex numbers. (a) 3 − 4i, (b) −2 + i, (c)
1 1 + i √ , (d) 3 7 − i −4 − 3i (e) 5(cos π/3 + i sin π/3), (f) cos 2π/3 − sin 2π/3

Answer 1

(a) 3 − 4i

$z=3-4i$ $|z|=\sqrt{(3)^2+(-4)^2}=5$ $Arg(z)=\arctan(-{4\over 3})=-\arctan({4\over 3})$

(b) −2 + i

$z=-2+i$ $|z|=\sqrt{(-2)^2+(1)^2}=\sqrt{5}$ $Arg(z)=\pi-\arctan({1\over 2})$

(c) 1/(1 + i√3)

$z={1\over 1+i\sqrt{3}}={1-i\sqrt{3}\over 4}$ $|z|=\sqrt{({1\over 4})^2+(-{\sqrt{3}\over 4})^2}={1\over 2}$ $Arg(z)=-{\pi\over 3}$

(d) (7 − i)/(−4 − 3i)

$z={7-i\over -4-3i}={1\over 25}(7-i)(-4+3i)=-1+i$ $|z|=\sqrt{(-1)^2+(1)^2}=\sqrt{2}$ $Arg(z)=\pi-{\pi\over 4}={3\pi\over 4}$

(e) 5(cos π/3 + isin π/3)

$z=5(\cos({\pi\over 3})+i\sin({\pi\over 3}))$ $|z|=5$ $Arg(z)={\pi\over 3}$

(f) cos 2π/3 −sin 2π/3.

$z=\cos({2\pi\over 3})-\sin({2\pi\over 3})=-{1\over 2}-{\sqrt{3}\over 2}$ $|z|={1+\sqrt{3}\over 2}$ $Arg(z)=\pi$