Question

1. Write

(2−i) (4−3i)2

3+2i

(4)

in the form a+bi, where a,b ∈ R.

Answer 1

$$\dfrac{(2−i) (4−3i)^2}{3+2i}=\dfrac{(2−i) (16-24i-9)(3-2i)}{(3)^2+(2)^2}$$

$$=\dfrac{(6-4i-3i-2) (7-24i)}{13}$$

$$=\dfrac{(4-7i) (7-24i)}{13}=\dfrac{28-96i-49i-168}{13}$$

$$=-\dfrac{140}{13}-\dfrac{145}{13}i$$