How do you write the following quotient in standard form #(6-7i)/i#?

1 Answer
Oct 6, 2016

Answer:

I found: #-7-6i#

Explanation:

You need to get rid of the #i# at the denominator first. To do that you can multiply and divide by the complex conjugate of the denominator;
given a complex number in the form #a+ib# the complex conjugate will be #a-ib# (change the sigh of the imaginary part). In your case you have #0+i# and the complex conjugate will be #0-i#.
So we have:
#(6-7i)/(0+i)*(0-i)/(0-i)=#
multiplying and remembering that #i^2=-1# we get:
#(-6i+7i^2)/(-i^2)=(-7-6i)/1=-7-6i# in standard form.