Question

Convert the following in the polar form: (i) , (ii)

Answer 1

(i)

The given expression for z is,

z= 1+7i ( 2−i ) 2

Simplify the above expression.

z= 1+7i ( 2−i ) 2 = 1+7i 4+ i 2 −4i = 1+7i 4−1−4i Multiplying the numerator and denominator by (3 + 4i), z= 1+7i 3−4i × 3+4i 3+4i

= 3+4i+21i+28 i 2 3 2 − ( 4i ) 2 = 3+4i+21i−28 3 2 + 4 2 = −25+25i 25 =−1+i

Let rcosθ=−1 and rsinθ=1

Squaring and adding the above expression,

r 2 ( cos 2 θ+ sin 2 θ )=1+1 r 2 ( cos 2 θ+ sin 2 θ )=2 r 2 =2 r= 2

Solving for θ

2 cosθ=−1,  2 sinθ=1 cosθ=− 1 2 , sinθ= 1 2

Thus, the value of θ , θ lies in second quadrant.

θ=π− π 4 = 3π 4

z=rcosθ+irsinθ = 2 cos 3π 4 +i 2 sin 3π 4 = 2 ( cos 3π 4 +isin 3π 4 )

Thus, the polar representation is 2 ( cos 3π 4 +isin 3π 4 ) .

(ii)

The given expression for z is,

z= 1+3i 1−2i

Multiplying the numerator and denominator by 1+2i

= 1+3i 1−2i × 1+2i 1+2i = 1+2i+3i−6 ( 1 ) 2 + ( −2i ) 2 = −5+5i 1+4 = −5+5i 5 =−1+i

Let rcosθ=−1, rsinθ=1

Squaring and adding the above expression,

r 2 ( cos 2 θ+ sin 2 θ )=1+1 r 2 ( cos 2 θ+ sin 2 θ )=2 r 2 =2 r= 2

Solving for θ

2 cosθ=−1,  2 sinθ=1 cosθ=− 1 2 , sinθ= 1 2

The value of θ , θ lies in second quadrant.

θ=π− π 4 = 3π 4

The expression for z is given by,

z=rcosθ+irsinθ = 2 cos 3π 4 +i 2 sin 3π 4 = 2 ( cos 3π 4 +isin 3π 4 )

Thus, the polar representation is 2 ( cos 3π 4 +isin 3π 4 ) .