Rationalize the denominator of 10 over quantity of 8 minus 5i

Question

Rationalize the denominator of  10 over quantity of 8 minus 5i

anonymous · Answer

multiply top and bottom by the conjuage of the denominator
the conjugate of $a+bi$ is $a-bi$ and this works because 
$$(a+bi)(a-bi)=a^2+b^2$$ a real number 
in your case 

$$\frac{10}{8-5i}=\frac{10}{8-5i}\times \frac{8+5i}{8+5i}$$
$$=\frac{10(8-5i)}{8^2+5^2}$$ simplify from there

anonymous · Answer

is it 80-50i over 69?

anonymous · Answer

i mean 89

one098 · Answer

@sammixboo I got 80-50i/89 but that isnt one of the choices..

sammixboo · Answer

Whoa... Old question :P

one098 · Answer

lol yeah sorry

sammixboo · Answer

Okay let me try in a few minutes. My mom just called me in to eat

one098 · Answer

ok

one098 · Answer

@Jhannybean

jhannybean · Answer

@satellite73 you cleared up my confusion about the technicalities involving conjugates of complex numbers, lol. thank you.

one098 · Answer

but I'm still as confused as always! .-.

jhannybean · Answer

How @satellite73 solved it was to simplify the fraction with the imaginary number at the bottom by multiplying by the conjugate, that is the opposite of the function. When he did this, he got : $8^2 +40i -40i -25i^2$ 

We know that $i^2 = -1$ so substituting that into the function for i^2 he simpplified it to $8^2 +5^2$

jhannybean · Answer

All the  OP has to do is expand the numerator.

one098 · Answer

I got 80-50i/89 but again, that isn't one of the choices...

jhannybean · Answer

Then their problem might be wrong.

one098 · Answer

:/ ok then