MTH632 Practice Questions Solution Lecture 3

z1 = – 8 + i, |z1| = (-8)2+12 = 65 = 8.06

z2 = –3 + 8i, |z2| = (-3)2+82 = 73 = 8.5

z1 + z2 = (– 8 + i ) + (–3 + 8i) = -11 + 9i

|z1+z 2| = (-11)2+92 = 202 = 14.12

Clearly |z1+z 2| < |z1| + |z2|

14.12 < 8.06+8.5

14.12 < 16.56

Q6. Find the Polar representation of the complex numbers z = – 7 + 5i.

Solution:

We represent z = ( x, y ) in (r, θ)in polar form.

For z = x + i y, r = | z | = (-7)2+52 =

θ = tan-1( y x ) = tan-1( -7 ) = -tan-1( 7 )

z = r(cos θ +isin θ)

z = 74 (cos (-tan-1( 7 ))+isin (-tan-1(

7 )))

Q7. Find the Polar representation of the complex numbers z = ( 0, 11 ) = 11i.

Solution:

We represent z = ( x, y ) in (r, θ)in polar form.

For z = x + i y, r = | z | = (0)2+112 = 121 = 11

θ = tan-1( y x ) = tan-1( 0 ) = ∞, so θ = π

z = r(cos θ +isin θ)