How do you divide #(-5-i)/(-10i)#?

1 Answer
Sep 12, 2016

Answer:

I found: #0.5-0.5i#

Explanation:

Here you have a division between complex numbers that normally wouldn't be immediately possible or at least will not give a complex number in standard form: #a+ib#.

To solve the problem we need to get rid of #i# in the denominator!
To do that we multiply and divide by the complex conjugate of the denominator .

If you have a complex number, say, #3+4i# the complex conjugate is obtained changing the sign of the immaginary part as: #3color(red)(-)4i#.

In our case the denominator is:
#0-10i#

so the complex conjugate will be: #0+10i#.

Let us multiply and divide:

#(-5-5i)/(0-10i)*(0+10i)/(0+10i)=#

After this (even if it looks more difficult) we do the multiplications:
#=(-50i-50i^2)/(-100i^2)=#

remember that # i^2=-1# so:

#=(50-50i)/(100)=#

Now, the good thing is that we got rid of #i# in the denominator AND we can rearrange separating into fractions as:
#=50/100-50/100i=0.5-0.5i=1/2-1/2i#
which is in the form #a+ib#!!!