Answer to Question #117714 in Algebra for Julia Oduro

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 √3 (d) (7 − i)/(-4 - 3i) (e) 5(cos π/3 + isin π/3) (f) cos 2π/3 − sin 2π/3.

Answer 1

The formula for modulus (r) and principal argument ($\theta$) of a complex number (z = x + iy) is :

r = $\mid$z$\mid$ = $\sqrt{\smash[b]{x² + y²}}$

$\theta$ = arctan(y/x)

(a) z = 3 - 4i

Modulus, r = $\sqrt{\smash[b]{3² + (-4)²}}$

r = $\sqrt{\smash[b]{9 + 16}}$ = $\sqrt{\smash[b]{25}}$

$\therefore$ r = 5

Principal argument, $\theta$ = arctan(-4/3)

$\therefore$ $\theta$ = -53.13$\degree$

(b) z = -2 + i

Modulus, r = $\sqrt{\smash[b]{(-2)² + (1)²}}$

r = $\sqrt{\smash[b]{4 + 1}}$

$\therefore$ r = $\sqrt{\smash[b]{5}}$

Using arctan(1/(-2))$= -26.56\degree$

Since the given complex number lies in the second quadrant,

Therefore, the principal argument is $\theta = (- 26.56 + 180)\degree$

$\therefore \theta = 153.44\degree$

(c) z = 1/(1 + i$\sqrt{\smash[b]{3}}$ )

$z = (1 - i\sqrt{\smash[b]{3}})/((1 + i\sqrt{\smash[b]{3}})(1 - i\sqrt{\smash[b]{3}}))$

$z = (1 - i\sqrt{\smash[b]{3}})/(1² - (i\sqrt{\smash[b]{3}})²)$

$z = (1 - i\sqrt{\smash[b]{3}})/(1 + 9)$

$z = (1 - i\sqrt{\smash[b]{3}})/10$

$z = (1/10) - i(\sqrt{\smash[b]{3}}/10)$

Modulus, r = $\sqrt{\smash[b]{(1/10)² + (\sqrt{\smash[b]{3}}/10)²}}$

r = $\sqrt{\smash[b]{(1/100) + (3/100)}}$

r = $\sqrt{\smash[b]{(4/100)}}$

$\therefore$ r = (2/10)

Principal argument, θ = arctan$( -(\sqrt{\smash[b]{3}}/10)/(1/10)$

$\theta = arctan(- \sqrt{\smash[b]{3}})$

$\theta = - ( \pi / 3)$

$\therefore$ $\theta = - 60\degree$

(d) z = (7 − i)/(-4 - 3i)

z = ((7 - i)(-4 + 3i))/((-4 - 3i)(-4 + 3i))

z = (-28 + 21i +4i + 3)/((-4)² - (3i)²)

z = (-25 + 25i)/(16 + 9)

z = (-25 + 25i)/25

z = -1 + i

Modulus, r = $\sqrt{\smash[b]{(-1)² + 1²}}$

r = $\sqrt{\smash[b]{1 + 1}}$

$\therefore r= \sqrt{\smash[b]{2}}$

Using arctan(1/(-1)) = arctan(-1)= -45$\degree$ .

Since the given complex number lies in the second quadrant,

Therefore, the principal argument is $\theta = (-45 + 180)°$

$\therefore\theta= 135°$

(e) z = $5(cos(\pi/3) +isin(\pi/3))$

z= $5cos(\pi/3) +i5sin(\pi/3)$

Modulus, r = $\sqrt{\smash[b]{(5cos(π/3))² + (5sin(π/3))²}}$

$r = \sqrt{\smash[b]{5²(cos²(π/3) + sin²(π/3))}}$

$r = 5 \sqrt{\smash[b]{1}}$

$\therefore$ r = 5

Principal argument, θ = arctan(5sin(π/3)/5cos(π/3))

θ = arctan(tan(π/3))

θ = $\pi/3$

$\therefore\theta$ = 60$\degree$

(f) z = cos 2π/3 − sin 2π/3

z = (-1/2) - ($\sqrt{\smash[b]{3}}/2$ )

$z = - (1 + \sqrt{\smash[b]{3}})/2$

Modulus, r = $|-(1+\sqrt{3})/2|=(1+\sqrt{3})/2$

$\therefore$ $r =(1+\sqrt{3})/2$

Since the given complex number is negative real number,

Therefore, the principal argument is $\theta=180°$

$\therefore\theta= 180°$