Question #117761
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.
1
Expert's answer
2020-05-28T18:28:16-0400

The formula for modulus and principle argument are as follows

z=x2+y2| z | = \sqrt{\smash[b]{x^2 + y^2}}



θ=arctan(y/x)\theta = arctan(y/x)


A) 3 - 4i


34i=(3)2+(4)2| 3 - 4i | = \sqrt{\smash[b]{(3)^2+ (-4)^2}}

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

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

r = 5


θ=arctan(4/3)\theta = arctan(-4/3)

θ\theta = - arctan(4/3)

θ\theta = - 53.13°



B) -2 + i


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

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

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


arctan(1/2)\arctan(1/-2)

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

Since the given complex number lies in the second quadrant.

Therefore, θ=(26.56+180)°\theta =( -26.56 + 180)°

θ=153.44°\theta = 153.44°


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


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

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

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

r=2/10r = 2/10


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

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

θ=(π/3)\theta = - ( \pi / 3)

θ=60°\theta = - 60°



D)z=(7i)/(43i)z = (7 - i)/(-4 - 3i)

z=((7i)(4+3i))/((43i)(4+3i))z = ((7 - i)(-4 + 3i))/((-4 - 3i)(-4 + 3i))

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

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

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

z=1+iz = -1 + i


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

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

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


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

=arctan(1)=\arctan(-1)

=45°= -45\degree

Since the given complex number lies in the second quadrant,

Therefore, θ=(45+180)°\theta=(-45 + 180)°

θ=135°\theta = 135°



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

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


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

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

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

r=5r = 5


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

θ=arctan(tan(π/3))\theta = arctan(tan(\pi/3))

θ=π/3=60°\theta =\pi/3 = 60\degree


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

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


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

Since the given complex number is negative real number,

θ=180°\theta =180°


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS