Question

Pre-calculus. PLEASE HELP! MEDAL! @Directrix @SolomonZelman @e.mccormick

Answer 1

you can use pascals triangle to expand it (or wolfram)

or first find what \(\large\color{slate}{ (-2-2i)^2 }\) is, and then raise that to the second power.

Answer 2

If I do that, I get -64. Which isn't an answer choice. This has to do with polar equation, complex numbers, etc.

Answer 3

you say isn't an answer choice?

Answer 4

oops, sorry it is an answer choice.

Answer 5

lo, sorry looking at the wrong problem.

Answer 6

I doubt it would be -64 though

Answer 7

Stupid me. :(
Anyway, thanks. :)

Answer 8

Yeah, I feel like it should be a or c, but I don't know how to get that

Answer 9

what I would do for convenience.

\(\large\color{slate}{ (-2-2i)^4 }\)

\(\large\color{slate}{ ((-1)(2+2i))^4 }\)

\(\large\color{slate}{ (-1)^4(2+2i)^4 }\)

\(\large\color{slate}{ (2+2i)^4 }\)

Answer 10

\(\large\color{slate}{ (2+2i)^4 }\)

\(\large\color{slate}{ \left(~\color{red}{(2+2i)^2}\right)^2 }\)

applying `(a+b)^2=(a^2+2ab+b^2)` to the red part

\(\large\color{slate}{ \left(~\color{red}{2^2+2(2)(2i)+(2i)^2}\right)^2 }\)

\(\large\color{slate}{ \left(~\color{red}{4+8i+(-4)}\right)^2 }\)

\(\large\color{slate}{ \left(~\color{red}{8i}\right)^2 }\)

Answer 11

yes, your answer was correct actually

Answer 12

I just was thinking that there will always be a middle term with i's and a constant near, but the constant in last and first term cancel each other