Question

.

Answer 1

@FlowerOfLife, this is already simplified. Do you need to convert to polar coordinates?

Answer 2

multiply numerator and denominator by 3+2i

Answer 3

Is the problem
$$
4/3 - 2i
$$
Or
$$
\cfrac{4}{3 - 2i}
$$

Answer 4

@FlowerOfLife recall the "conjugates" from the previous posting?

use the conjugate of the denominator and multiply both, top and bottom by it

Answer 5

what would be the conjugate of " 3 - 2i "?

Answer 6

sorry, but the connection is a bit lagged, thus.... yeas. is 3+2i so, one sec

Answer 7

\(\bf \cfrac{4}{3-2i}\qquad \qquad \textit{conjugate of }3-2i\textit{ is }3+2i\quad thus\\ \quad \\
\cfrac{4}{3-2i}\times \cfrac{3+2i}{3+2i}\implies \cfrac{4(3+2i)}{(3-2i)(3+2i)}\\ \quad \\
\textit{keep in mind that }\qquad \color{blue}{(a-b)(a+b) = (a^2-b^2)}\\ \quad \\
\cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{12-8i}{3^2-(2i)^2}\\ \quad \\
\color{blue}{i^2 \implies \sqrt{-1}\times \sqrt{-1}\implies \sqrt{(-1)^2}\implies -1}\\ \quad \\
\cfrac{12-8i}{3^2-(2i)^2}\implies \cfrac{12-8i}{3^2-2^2\cdot i^2}\implies \cfrac{12-8i}{3^2-2^2\cdot -1}\implies\cfrac{12-8i}{9+4}\)

Answer 8

hmmm... I have ... a - up there... has to be a +... .lemme fix that quick

Answer 9

\(\bf \cfrac{4}{3-2i}\qquad \qquad \textit{conjugate of }3-2i\textit{ is }3+2i\quad thus\\ \quad \\
\cfrac{4}{3-2i}\times \cfrac{3+2i}{3+2i}\implies \cfrac{4(3+2i)}{(3-2i)(3+2i)}\\ \quad \\
\textit{keep in mind that }\qquad \color{blue}{(a-b)(a+b) = (a^2-b^2)}\\ \quad \\
\cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{12+8i}{3^2-(2i)^2}\\ \quad \\
\color{blue}{i^2 \implies \sqrt{-1}\times \sqrt{-1}\implies \sqrt{(-1)^2}\implies -1}\\ \quad \\
\cfrac{12+8i}{3^2-(2i)^2}\implies \cfrac{12+8i}{3^2-2^2\cdot i^2}\implies \cfrac{12+8i}{3^2-2^2\cdot -1}\implies\cfrac{12+8i}{9+4}
\)