Question

Simplify and write the answer in standard form 4+3i/ 5+2i

Answer 1

Hi there,
\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

We want to multiply the top and bottom by the conjugate of our denominator,
\[\Large\bf\sf \frac{4+3\mathcal i}{5+2\mathcal i}\color{royalblue}{\left(\frac{5-2\mathcal i}{5-2\mathcal i}\right)}\]

Answer 3

Remember your rule for multiplying complex conjugates?\[\Large\bf\sf (a-b\mathcal i)(a+b\mathcal i)\quad=\quad a^2+b^2\]

Answer 4

So we get,\[\Large\bf\sf =\quad \frac{(4+3\mathcal i)(5-2\mathcal i)}{5^2+2^2}\]

Answer 5

Do you understand how to proceed from there?

Answer 6

can you continue please

Answer 7

Nooo you gotta do some of the work here silly!! :)
Do you remember how to multiply brackets out?
Does FOIL sound familiar maybe?

Answer 8

i know how to foil so my top part will look like (4+3i)(4-3i)(5+2i)(5-2i)

Answer 9

No, your top is,\[\Large\bf\sf (4+3\mathcal i)(5-2\mathcal i)\]And you need to FOIL that out.

Answer 10

but i thought every complex number came with its other pair(+or-)

Answer 11

No.. hmm not sure where you're getting that from :D

All complex `roots` come in pairs, yes.
So when you're solving an equation for x or something, then yes you'll always get the conjugate along with your complex root.

But not when we're just straight up dealing with complex numbers.

Answer 12

ooooohh sorry i see now so my top answer is -6i^2+7i+20

Answer 13

all over 29

Answer 14

Mmmm ok looks good so far!

Answer 15

Simplify the i^2 term.
You should be able to combine it with something else after that.

Answer 16

so its 7i+26

Answer 17

Ok looks good so far :)
\[\Large\bf\sf \frac{26+7\mathcal i}{29}\]

Answer 18

Now we want to get it into standard form.

Answer 19

i thought tha was it?

Answer 20

We want to split this into two fractions, the real and imaginary parts.

Answer 21

"but i thought every complex number came with its other pair(+or-)" this is correct when rationalizing
Ex. 2i/2i-5
2i(2i+5)/(2i-5)(2i+5)

Answer 22

I don't know what this is called, but I like to call it the jelly bean method :)