Write (-1+3i)^(4) in simplest a+bi form. Answer:

Calculate First Square: To solve (-1+3i)^(4), we will first calculate the square of (-1+3i) and then square the result again to get the fourth power. (-1+3i) * (-1+3i) = (-1)^2 + 2*(-1)*3i + (3i)^2 = 1 - 6i - 9(i^2) Since i^2 = -1, we replace i^2 with -1. = 1 - 6i - 9(-1) = 1 - 6i + 9 = 10 - 6iReplace i^2 with -1: Now we need to square the result (10 - 6i) to get the fourth power of (-1+3i). (10 - 6i) * (10 - 6i) = 10^2 - 2*10*6i + (6i)^2 = 100 - 120i + 36(i^2) Again, we replace i^2 with -1. = 100 - 120i + 36(-1) = 100 - 120i - 36 = 64 - 120i

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Write
(-1+3i)^(4) in simplest
a+bi form.
Answer:

Write $(-1+3 i)^{4}$ in simplest $a+b i$ form. $\newline$ Answer:

View step-by-step help

Home

Math Problems

Precalculus

Find the roots of factored polynomials

Full solution

Q. Write

(-1+3 i)^{4}

in simplest

a+b i

form.

\newline

Answer:

Calculate First Square: To solve $(-1+3i)^{4}$ , we will first calculate the square of $(-1+3i)$ and then square the result again to get the fourth power. $\newline$ (-1+3i) \times (-1+3i) = (-1)^2 + 2\times(-1)\times3i + (3i)^2$\newline= 1 - 6i - 9(i^2)$ $\newline$ Since $i^2 = -1$ , we replace $i^2$ with $-1$ . $\newline$ $= 1 - 6i - 9(-1)$ $\newline$ $= 1 - 6i + 9$ $\newline$ $= 10 - 6i$
Replace $i^2$ with $-1$ : Now we need to square the result $(10 - 6i)$ to get the fourth power of $(-1+3i)$ .
$(10 - 6i) \times (10 - 6i) = 10^2 - 2\times10\times6i + (6i)^2$
$= 100 - 120i + 36(i^2)$
Again, we replace $i^2$ with $-1$ .
$= 100 - 120i + 36(-1)$
$= 100 - 120i - 36$
$-1$ $0$