help!!!!! a little lost

[latresa31s]

can you show me how to divide this?

-12i/(2-5i)

 

[Daniel_Feldman]

Multiply both the numerator and the denominator by 2+5i and simplify.

 

[latresa31s]

OK am I setting this up right?

(-12i)*(2+5i)= -24+-60i^2

(2+5i)*(2-5i)=4-25i^2

-24i + -60i(-1)/4-25(-1)

I am not sure if this is right could you let me know?

 

[pka]

Note that |2−5i|<SUP>2</SUP>=29.
(-12i)*(2+5i)= -24i−60i<SUP>2</SUP>=60−24i.
So -12i/(2-5i)=[60−24i]/29.

 

[latresa31s]

how did you get 29

ok

-12i
(2-5i)

= (-12i)(2+5i)
(2-5i)(2+5i)

= 24i-60i^2
-21i

Ok this is far as I got to.

 

[pka]

The correct answer is [60−24i]/29.
You have real problems with signs!

 

[latresa31s]

Pka sorry to be such a problem. but could you show me how this problem would be set up I would like to use it as an example for my other problems.

 

[pka]

Note that (2−5i)(2+5i)=(4−25i<SUP>2</SUP>)=(4−25(−1))=29.
(-12i)*(2+5i)= −24i−60i<SUP>2</SUP>=60−24i.
So -12i/(2-5i)=[60−24i]/29.

 

[latresa31s]

Thanks alot