Answer to Question #117368 in Algebra for jay

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 (θ) of a complex number (z = x + iy) is :

r = ∣z∣ = $\sqrt{\smash[b]{x² + y²}}$

θ = arctan(y/x)

(a) z = 3 - 4i

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

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

∴ r = 5

Principal argument, θ = arctan(-4/3)

∴ θ = -53.13°

(b) z = -2 + i

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

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

∴ r = $\sqrt{\smash[b]{5}}$

Principal argument, θ = arctan(1/(-2))

θ=−26.56°

Since the given complex number lies in the second quadrant,

Therefore, θ=(−26.56+180)°

∴θ=153.44°

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

​

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

∴ r = (2/10)

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

θ=arctan$(- \sqrt{\smash[b]{3}})$

θ=−(π/3)

∴ θ=−60°

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

∴r=$\sqrt{\smash[b]{2}}$

Principal argument, θ = arctan(1/(-1))

θ = arctan(-1)

θ = -45°

Since the given complex number lies in the second quadrant,

Therefore, θ=(−45+180)°

∴θ=135°

(e) z = 5(cos(π/3)+isin(π/3))

z= 5cos(π/3)+i5sin(π/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 $\sqrt{\smash[b]{1}}$

∴ r = 5

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

θ = arctan(tan(π/3))

θ = π/3

∴θ = 60°

(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 =$\sqrt{\smash[b]{(cos(2π/3) - sin(2π/3))² + 0}}$

$r = \sqrt{\smash[b]{(cos²(2π/3) + sin²(2π/3) - 2sin(2π/3)cos(2π/3)}}$

$r = \sqrt{\smash[b]{(1 - sin(4π/3))}}$

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

∴ r =$\sqrt{\smash[b]{(2 + \sqrt{\smash[b]{3}})/2) }}$

Principal argument, θ = arctan(0/(cos(2π/3) - sin(2π/3)))

θ=arctan(0)

Since the given complex number lies in the second quadrant,

Therefore, θ=(0+180)°

∴θ=180°