Answer to Question #117761 in Algebra for Asubonteng Isaac Adjei

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(cosπ/3 + isinπ/3), (f) cos2π/3−sin2π/3.

Answer 1

The formula for modulus and principle argument are as follows

$| z | = \sqrt{\smash[b]{x^2 + y^2}}$

$\theta = arctan(y/x)$

A) 3 - 4i

$| 3 - 4i | = \sqrt{\smash[b]{(3)^2+ (-4)^2}}$

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

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

r = 5

$\theta = arctan(-4/3)$

$\theta$ = - arctan(4/3)

$\theta$ = - 53.13°

B) -2 + i

$| -2 + i | = \sqrt{\smash[b]{(-2)^2 + (1)^2}}$

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

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

$\arctan(1/-2)$

$= - \arctan(1/2) = -26.56°$

Since the given complex number lies in the second quadrant.

Therefore, $\theta =( -26.56 + 180)°$

$\theta = 153.44°$

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

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

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

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

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

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

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

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

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

$r = 2/10$

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

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

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

$\theta = - 60°$

D)$z = (7 - i)/(-4 - 3i)$

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

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

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

$z = (-25 + 25i)/25$

$z = -1 + i$

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

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

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

$\arctan(1/(-1))$

$=\arctan(-1)$

$= -45\degree$

Since the given complex number lies in the second quadrant,

Therefore, $\theta=(-45 + 180)°$

$\theta = 135°$

E) $5(cos(\pi/3) +isin(\pi/3))$

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

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

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

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

$r = 5$

$\theta = arctan( 5sin(\pi/3) / 5cos(\pi/3) )$

$\theta = arctan(tan(\pi/3))$

$\theta =\pi/3 = 60\degree$

F)$z = cos(2\pi/3) - sin(2\pi/3))$

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

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

Since the given complex number is negative real number,

$\theta =180°$