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.
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Expert's answer
2020-05-28T18:17:52-0400
The formula for modulus (r) and principal argument (θ) of a complex number (z = x + iy) is :
r = ∣z∣ = x2+y2
θ = arctan(y/x)
(a) z = 3 - 4i
Modulus, r = 32+(−4)2
r = 9+16 = 25
∴ r = 5
Principal argument, θ = arctan(-4/3)
∴θ = -53.13°
(b) z = -2 + i
Modulus, r = (−2)2+(1)2
r = 4+1
∴ r = 5
Using arctan(1/(-2))=−26.56°
Since the given complex number lies in the second quadrant,
Therefore, the principal argument is θ=(−26.56+180)°
∴θ=153.44°
(c) z = 1/(1 + i3 )
z=(1−i3)/((1+i3)(1−i3))
z=(1−i3)/(12−(i3)2)
z=(1−i3)/(1+9)
z=(1−i3)/10
z=(1/10)−i(3/10)
Modulus, r = (1/10)2+(3/10)2
r = (1/100)+(3/100)
r = (4/100)
∴ r = (2/10)
Principal argument, θ = arctan(−(3/10)/(1/10)
θ=arctan(−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 = (−1)2+12
r = 1+1
∴r=2
Using arctan(1/(-1))= arctan(-1)= -45° .
Since the given complex number lies in the second quadrant,
Therefore, the principal argument is θ=(−45+180)°
∴θ=135°
(e) z = 5(cos(π/3)+isin(π/3))
z= 5cos(π/3)+i5sin(π/3)
Modulus, r = (5cos(π/3))2+(5sin(π/3))2
r=52(cos2(π/3)+sin2(π/3))
r=51
∴ 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) - (3/2 )
z=−(1+3)/2
Modulus, r = ∣−(1+3)/2∣=(1+3)/2
∴r=(1+3)/2
Since the given complex number is negative real number,
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