Question #117714
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.
1
Expert's answer
2020-05-28T18:17:52-0400

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


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

θ\theta = arctan(y/x)


(a) z = 3 - 4i

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

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

\therefore r = 5


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

\therefore θ\theta = -53.13°\degree

(b) z = -2 + i

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

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

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


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

Since the given complex number lies in the second quadrant,

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

θ=153.44°\therefore \theta = 153.44\degree


(c) z = 1/(1 + i3\sqrt{\smash[b]{3}} )

z=(1i3)/((1+i3)(1i3))z = (1 - i\sqrt{\smash[b]{3}})/((1 + i\sqrt{\smash[b]{3}})(1 - i\sqrt{\smash[b]{3}}))

z=(1i3)/(12(i3)2)z = (1 - i\sqrt{\smash[b]{3}})/(1² - (i\sqrt{\smash[b]{3}})²)

z=(1i3)/(1+9)z = (1 - i\sqrt{\smash[b]{3}})/(1 + 9)

z=(1i3)/10z = (1 - i\sqrt{\smash[b]{3}})/10

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


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

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

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

\therefore r = (2/10)


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

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

θ=(π/3)\theta = - ( \pi / 3)

\therefore θ=60°\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 = (1)2+12\sqrt{\smash[b]{(-1)² + 1²}}

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

r=2\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 θ=(45+180)°\theta = (-45 + 180)°

θ=135°\therefore\theta= 135°


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

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


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

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

r=51r = 5 \sqrt{\smash[b]{1}}

\therefore r = 5


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

θ = arctan(tan(π/3))

θ = π/3\pi/3

θ\therefore\theta = 60°\degree


(f) z = cos 2π/3 − sin 2π/3

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

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


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

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


Since the given complex number is negative real number,

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

θ=180°\therefore\theta= 180°


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