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.
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Expert's answer
2020-05-28T18:28:16-0400
The formula for modulus and principle argument are as follows
∣z∣=x2+y2
θ=arctan(y/x)
A) 3 - 4i
∣3−4i∣=(3)2+(−4)2
r = 9+16
r = 25
r = 5
θ=arctan(−4/3)
θ = - arctan(4/3)
θ = - 53.13°
B) -2 + i
∣−2+i∣=(−2)2+(1)2
r=4+1
r=5
arctan(1/−2)
=−arctan(1/2)=−26.56°
Since the given complex number lies in the second quadrant.
Therefore, 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)−i3/10
Modulus,r=(1/10)2+(3/10)2
r=(1/100)+(3/100
r=(4/100)
r=2/10
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
arctan(1/(−1))
=arctan(−1)
=−45°
Since the given complex number lies in the second quadrant,
Therefore, θ=(−45+180)°
θ=135°
E) 5(cos(π/3)+isin(π/3))
=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
θ=arctan(5sin(π/3)/5cos(π/3))
θ=arctan(tan(π/3))
θ=π/3=60°
F)z=cos(2π/3)−sin(2π/3))
z=(−1/2)−(3/2)=−(1+3)/2
Modulus,r=∣−(1+3)/2∣=(1+3)/2
Since the given complex number is negative real number,
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