Question

Question 5 Convert the following in the polar form:

(i) (1+7i)/(2-i)^2

(ii) (1+3i)/(1-2i)

Class X1 - Maths -Complex Numbers and Quadratic Equations Page 112

Answer 1

(i) (1 + 7i)/(2 - i)²
= (1 + 7i)/{(2)² +(i)²-2.2i}
=(1 + 7i)/(4 -1 - 4i) [ we know, i² -1 use it here]
= (1 + 7i)/(3 - 4i)
Multiply (3 + 4i) both sides,
= (1 + 7i)(3 + 4i)/(3-4i)(3+4i)
= { 1(3 + 4i) +7i(3 + 4i)}/{(3²-(4i)²}
= ( 3 + 4i + 21i -28 )/(9 +16)
= (25i - 25)/25
= i -1 or -1 + i

Let -1 + i =rcos∅ + irsin∅
rcos∅ = -1 -----(1)
rsin∅ = 1 ------(2)

square and add both eqns
r²( sin²∅ + cos²∅) = 1² + 1²
r = √2

divide (2)÷ (1)

tan∅ = 1 = tan(π/4)
∅ = π/4
but ∅ lies on 2nd quadrant .
arg(z) = π - ∅
= π - π/4
= 3π/4

hence , -1 + i in Polar form is
√2[ cos(3π/4) + isin(3π/4)]

(ii) similarly you can do work with it .
z = ( 1 + 3i)/(1-2i)
multiply (1 + 2i) both sides,
= (1 + 3i)(1 + 2i)/(1-2i)(1+2i)
= {1 +2i + 3i + 6i²}/(1² -(2i)²}
= { 1 + 5i - 6}/(1 +4)
= ( 5i - 5)/5
= -1 + i
It is same as above .
Hence, Polar form of given complex number is √2[cos(3π/4) +isin(3π/4)]