Simplify the expression the root of negative one all over the quantity of four minus three i minus the quantity of two plus five i.
Really don't get this please help.

Question

Simplify the expression the root of negative one all over the quantity of four minus three i minus the quantity of two plus five i. 
Really don't get this please help.

anonymous · Answer

$$\sqrt{-1} \over  \left( 4-3i \right) - \left( 2 +5i \right)$$

ganeshie8 · Answer

simplify the denominator first

ganeshie8 · Answer

combine like terms..

anonymous · Answer

$$\sqrt{-1} \over \left( 2-5i \right)$$

anonymous · Answer

Is that correct?

anonymous · Answer

@ganeshie8

ganeshie8 · Answer

try again

anonymous · Answer

is the denominator -8 +15i?

ganeshie8 · Answer

$\large    \sqrt{-1} \over  \left( 4-3i \right) - \left( 2 +5i \right)$

$\large    \sqrt{-1} \over   4-3i  - 2 - 5i $

$\large    \sqrt{-1} \over   2-8i  $

ganeshie8 · Answer

and since $\sqrt{-1} = i$, ur numerator   changes to  $i$

anonymous · Answer

oh okay I'm sorry and from there the numerator is i sqrt 1

ganeshie8 · Answer

$\large    \sqrt{-1} \over  \left( 4-3i \right) - \left( 2 +5i \right)$

$\large    \sqrt{-1} \over   4-3i  - 2 - 5i $

$\large    \sqrt{-1} \over   2-8i  $

$\large    i\over   2-8i  $

ganeshie8 · Answer

fine so far ?

anonymous · Answer

yes these are my choices though so I dont knwo where to go from there ):

ganeshie8 · Answer

you need to get rid of i's in the denominator and put this in standard form

rvc · Answer

i (2+8i)/4+8)
= 2i-8/12
=2(i-4)/12
=i-4/6
this is ur answer i suppose
@rosepedal1791

rvc · Answer

this is in continuation with ganesh's ans

anonymous · Answer

How can I do that? I'm a bit confused. and that isn't one of my choices, this is a really hard question. I learned about imaginary numbers but not to this extent.

rvc · Answer

hope u understand

rvc · Answer

just take the conjugate of denominator @rosepedal1791

anonymous · Answer

yep, take the the conjugate of denominator

ganeshie8 · Answer

let me show u in latex

anonymous · Answer

Okay, thank you. But there is no imaginary number in the denominator only the numerator in your answer. so i +4 / 6 ?

ganeshie8 · Answer

$\large    i\over   2-8i  $

multiply conjugate of denominator, top and bottom :

$\large    \frac{i}{ 2-8i }\times  \frac{2+8i}{2+8i} $

$\large    \frac{i(2+8i)}{ 2^2+8^2 } $

$\large    \frac{2i - 8}{ 68 } $

$\large    \frac{-4 + i}{ 34 } $

anonymous · Answer

Oh okay ! Thank you so much, I had no clue I had to multiply the top and bottom, never learned this before I have no clue why it's on my test. Thank you everyone. Im new so how do I give out medals lol

ganeshie8 · Answer

:) its kindof like rationalizing denominator...

ganeshie8 · Answer

nobody likes to leave radicals in the denominator

ganeshie8 · Answer

so we multiply wid the conjugate and get rid of them

ganeshie8 · Answer

similarly, we dont want "i"s in the denominator,
so we multiply wid conjugate and fix the denominator

anonymous · Answer

Rationalizing denominators, never got taught that. Teachers can be silly. But thank you I get what you are saying now. i over 2 - 8i didn't look right at first lol (:

ganeshie8 · Answer

good to hear that :)

rvc · Answer

keep it up!! @rosepedal1791