Question

Simplify the expression the root of negative sixteen all over the quantity of three minus three i plus the quantity of one minus two i..

Answer 1

I think the question is simplify

$\dfrac{\sqrt{-16}}{3 - 3i} + 1-2i$

$= \dfrac{4\sqrt{-1}}{3(1-i)}+ 1-2i$

$= \dfrac{4}{3}\dfrac{i}{1-i} \cdot \dfrac{1+i}{1+i} + 1-2i$

$= \dfrac{4}{3} \dfrac{i-1}{2} + 1-2i$

$= \dfrac{2}{3} i - \dfrac{2}{3} + 1-2i$

$= \dfrac{1}{3} - \dfrac{4}{3} i$

Answer: (1/3) - (4/3)i

I might be getting the question wrong. It might be

$\dfrac{\sqrt{-16}}{3 - 3i + (1-2i)}$

Let's do this one too.

$= \dfrac{4i}{4 - 5i} \cdot \dfrac{4+5i}{4+5i}$

$= \dfrac{16i - 20}{4^2 + 5^2}$

$=-\dfrac{20}{41} + \dfrac{16}{41} i$