Question

Find the modulus and argument of the complex number .

Answer 1

Let,

z= 1+2i 1−3i = 1+2i 1−3i × ( 1+3i ) ( 1+3i ) = 1+3i+2i+6 i 2 1 2 + 3 2 = 1+5i+6( −1 ) 1+9 = −5+5i 10

Simplifying further, we get

z= −5 10 + 5i 10 = −1 2 + 1 2 i

(1)

Let z=rcosθ+irsinθ (2)

Comparing equation (1) and (2), we get

rcosθ= −1 2 , rsinθ= 1 2

Squaring and adding the above equations,

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

Solving for θ .

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

The value of θ , θ lies in second quadrant.

θ=π− π 4 = 3π 4

Therefore, the value of modulus and argument of the given complex number are 1 2 and 3π 4 respectively.