Question

2i over 3+7i??

Answer 1

The denominator is 3+7i

The conjugate of 3+7i is 3-7i

Answer 2

Multiply top and bottom of the fraction by 3-7i to rationalize the denominator

Answer 3

What do you get when you do that?

Answer 4

so the 7i would cancel right

Answer 5

what do you get when you expand out (3+7i)(3-7i)

if you're not sure, use the rule (a+b)(a-b) = a^2 - b^2

Answer 6

58?

Answer 7

good

Answer 8

so is that my answer?

Answer 9

so far, we have this

\[\Large \frac{2i}{3+7i}\]


\[\Large \frac{2i(3-7i)}{(3+7i)(3-7i)}\]


\[\Large \frac{2i(3-7i)}{(3)^2-(7i)^2}\]


\[\Large \frac{2i(3-7i)}{9-49i^2}\]


\[\Large \frac{2i(3-7i)}{9-49(-1)}\]


\[\Large \frac{2i(3-7i)}{9+49}\]


\[\Large \frac{2i(3-7i)}{58}\]

Answer 10

Then we distribute in the numerator to get

2i(3-7i)

6i-14i^2

6i-14(-1)

6i+14

14+6i

Answer 11

Thank you so doing that .. :) I can see how you got it

Answer 12

yw

Answer 13

oh wait, I forgot to reduce, one sec

Answer 14

Here is the full step by step picture


\[\Large \frac{2i}{3+7i}\]


\[\Large \frac{2i(3-7i)}{(3+7i)(3-7i)}\]


\[\Large \frac{2i(3-7i)}{(3)^2-(7i)^2}\]


\[\Large \frac{2i(3-7i)}{9-49i^2}\]


\[\Large \frac{2i(3-7i)}{9-49(-1)}\]


\[\Large \frac{2i(3-7i)}{9+49}\]


\[\Large \frac{2i(3-7i)}{58}\]


\[\Large \frac{6i-14i^2}{58}\]


\[\Large \frac{6i-14(-1)}{58}\]


\[\Large \frac{6i+14}{58}\]


\[\Large \frac{14+6i}{58}\]


\[\Large \frac{2(7+3i)}{58}\]


\[\Large \frac{7+3i}{29}\]

Answer 15

you were correct of course. Now how did you get that work to do that? I need that. haha

Answer 16

you mean how did I format the special math type?

Answer 17

there's an equation button below the text box where you can insert math symbols like fractions, square roots, etc

Answer 18

the equation button?

Answer 19

sqrt-50? is that 5isqrt2?

Answer 20

yes, next to the draw and attach file buttons

Answer 21

and yes, \[\Large \sqrt{-50} = 5i\sqrt{2}\]

Answer 22

yay!

Answer 23

nice work

Answer 24

okay this one is getting me
\[(6+\sqrt-3)^2\]

Answer 25

how far did you get?

Answer 26

I got 6+isqrt3^2

Answer 27

that is incorrect

Answer 28

okay

Answer 29

\[\Large (6+\sqrt{-3})^2\]

is the same as

\[\Large (6+\sqrt{-3})(6+\sqrt{-3})\]

Answer 30

to expand that out, I would first make a 2x2 table like this

Answer 31

then write the terms of each binomial like so