9+2i divided by 1-4i?

Question

9+2i divided by 1-4i
YOU HAVE TO SMART TO GET THIS

Answer 1

(9 + 2i) / (1 - 4i)

What you have to do is multiply the top and bottom by the conjugate of the bottom. Note that the result of multiplying a binomial by its conjugate yields a difference of squares. That is, if we have (a - b), and we multiply it by (a + b), we get a^2 - b^2. We WANT to do this because we want to get rid of the i on the bottom.

Multiplying top and bottom by (1 + 4i), we obtain

[(9 + 2i) (1 + 4i)] / [(1 - 4i) (1 + 4i)]

[9 + 36i + 2i + 8i^2] / [1 - 16i^2]

Remember that i^2 = -1. Grouping like terms and simplifying, we get

[9 + 38i + 8(-1)] / [1 - 16(-1)]

Simplifying some more,

[9 + 38i - 8] / [1 + 16]
[1 + 38i] / [17]

Now, we put each term over 17 because we want our complex number to be in the form a + bi

(1/17) + (38/17)i

Answer 2

(9+2i)/(1-4i) = (9+2i)(1+4i) / (1-4i)(1+4i) = (1+38i)/17
I've multiplied the top and bottom by the conjugate of the denominator and simplified.
I digress with your presumed definition of smart by the way...

Answer 3

Answer = 9-1/2i

Answer 4

(9+2i) / (1-4i) =
(9+2i)(1+4i) / (1-4i)(1+4i) =
(9+38i+8i^2) / (1-16i^2) =
(9+38i-8) / (1+16) =
(1+38i) / 17

Answer 5

9 + ((2 * i) divided by 1) - (4 * i) = 9 - 2 i

I WAS REALLY SMART AND WENT TO SEARCH IT ON GOOGLE MUAHAHAHAHAHAHAHAHAHAHA!!

Answer 6

Not divisible.

Answer 7

let me guess..your mommy didn't do your homework tonight..lol

Answer 8

8i