How do you simplify #(3-5i)/i#?

1 Answer
Sep 28, 2016

Answer:

I got: #-5-3i#

Explanation:

We normally tend to solve to get to the standard form: #a+ib#
For this reason we need to get rid of the imagianary unit in the denominator.

To do this we can multiply AND divide by the COMPLEX CONJUGATE of the denominator (i.e., the same complex number BUT with different sign for the imaginary part).

Here the denominator is: #0+i#

The complex conjugate will then be #0-i#

So we get:
#(3-5i)/i*(0-i)/(0-i)=(-3i+5i^2)/(-i^2)=#
remember that #i^2=-1# so you get:
#(-5-3i)/1=-5-3i#
which is in the generally required form #a+ib#