I cant take it anymore. Can someone please walk me through on how to simplify 8 over 2 plus 2i.

Question

anonymous · Answer

Multiply both the top and bottom by 2-2i so they cancel out

anonymous · Answer

By the way, if you hear the word "conjugate" in class, what it means is to change the sign in the middle of a binomial or complex number. 2-2i is the conjugate of 2+2i.

jim_thompson5910 · Answer

Multiply top and bottom by 2-2i to get...


$$\Large \frac{8}{2+2i}$$


$$\Large \frac{8(2-2i)}{(2+2i)(2-2i)}$$


$$\Large \frac{8(2-2i)}{(2)^2-(2i)^2}$$


$$\Large \frac{8(2-2i)}{4-4i^2}$$


$$\Large \frac{8(2-2i)}{4-4(-1)}$$


$$\Large \frac{8(2-2i)}{4+4}$$


$$\Large \frac{8(2-2i)}{8}$$


$$\Large \frac{\cancel{8}(2-2i)}{\cancel{8}}$$


$$\Large 2-2i$$



Therefore,


$$\Large \frac{8}{2+2i} = 2-2i$$



Notes:

* We multiply top and bottom by the denominators conjugate. The conjugate of a+bi is a-bi

* In step 3, I'm using the difference of squares rule (a+b)(a-b) = a^2 - b^2

* $\Large i = \sqrt{-1}$, so $\Large i^2 = -1$

anonymous · Answer

oh wow. thank you very much for showing me the process.i can use this to do other problems as well. thank you

jim_thompson5910 · Answer

you're welcome