Question

(-2√5 - 3i)^2Write your answer in the form a + bi. Simplify all radicals.

Answer 1

First, we need to solve it as:

`(-2√(5)-3i)^2=(-2√(5))^2-2\cdot(-2√(5))\cdot3i+(3i)^2`

Because:

`(a-b)^2=a^2-2\cdot a\cdot b+b^2`

Additionally:

`i^2=-1`

So, the initial expression is equal to:

`\begin{gathered} (-2√(5)-3i)^2=(-2√(5))^2-(2\cdot(-2√(5))\cdot3i)+(3i)^2 \\ (-2√(5)-3i)^2=(-2)^2\cdot(√(5))^2+12√(5)i+(3^2\cdot i^2) \\ (-2√(5)-3i)^2=4\cdot5+12√(5)i+9\cdot(-1) \\ (-2√(5)-3i)^2=20+12√(5)i-9 \\ (-2√(5)-3i)^2=11+12√(5)i \end{gathered}`

Answer: 11 + (12√5)i