Like what you saw?
Create FREE Account and:
 Watch all FREE content in 21 subjects(388 videos for 23 hours)
 FREE advice on how to get better grades at school from an expert
 Attend and watch FREE live webinar on useful topics
Dividing Complex Numbers  Concept
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Fractions with negative roots in the denominator or with i in the denominator must be rationalized (since i represents a square root). When dividing complex numbers with negative roots, simplify in terms of imaginary numbers and then multiply the numerator and denominator by i. When a binomial is in the denominator, rewrite using i and then multiply the numerator and denominator by the conjugate.
Dividing by a complex number or a number involving i. So whenever we're dividing by a number that involves i, what we have to do is rationalize the denominator. Remember that i is equal to the square root of 1 and we're not allowed to have square roots in the denominator so we have to get rid of it. Okay. So we're going to go back to a problem that we already know how to do. 6 over root 8. So whenever we're dealing with a problem like this we have to rationalize the denominator. Get rid of that square root. So there's two ways of doing it. You could either multilply by root 8 over root 8 and get rid of that or what I tend to do is I like dealing with smaller numbers so if I can I try to simplify that denominator first.
I know that 8 is the same thing as 4 times 2. Let's do a different color so we can see it. So what this is actually really equal to is 6 over 2 root 2. So now instead of having them multiply by root 8, I still need to get rid of a radical but I can multiply by root 2 instead. So we multiply by root 2 and then [IB] to get to the square root and square the 2 in the top as well. Okay.
Before I multiply that through I can see that I can simplify this. We have 6 over 2. This is going to cancel leaving me with 3. Okay? So we now have 3 root 2 in the numerator and then we have the 2 is gone away. So we have root 2 over times root 2. Square roots. When you multiply them together they just cancel each other out leaving us with what's inside which is 2. So what we ended up with is 3 root 2 over 2. Okay?
So same exact idea when we are dealing with imaginary numbers, numbers involving i. So right here we have 5 over square root of 9. The first thing I want to do is to simplify that denominator radical, okay? This is square root of 9 is 3. So this is going to be 3i in the denominator. Okay? So rewriting this we have 5 over 3i. The 3 isn't presenting a problem, so we can leave it as this but what we really want to do is get rid of that i. Remember that i times i, i squared is 1. So if we multiply this by i ihn the denominator, we'll get i squared, 1. Our square root is gone. We have to multiply by 1, so we need an i in the top as well. Simplifying this out we got 5i in the numerator over 3i squared in the denominator. i squared, 1 so this just becomes 5i over 3 okay?
So just like we did with normal radicals, whenever we're dealing with the radical of a negative we still have to get rid of it. I find it best to simplify my numbers so I deal with smaller things. But the main problem is is to get rid of that square root in the denominator.
Please enter your name.
Are you sure you want to delete this comment?
Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”
BRIGHTSTORM IS A REVOLUTION !!!”
because of you i ve got a 100/100 in my test thanks”
Concept (1)
Sample Problems (7)
Need help with a problem?
Watch expert teachers solve similar problems.

Dividing Complex Numbers
Problem 1 3,908 viewsSimplify:
√20 √72 
Dividing Complex Numbers
Problem 2 3,467 viewsSimplify:
1 + √4 4 + √9 
Dividing Complex Numbers
Problem 3 608 views 
Dividing Complex Numbers
Problem 4 556 views 
Dividing Complex Numbers
Problem 5 618 views 
Dividing Complex Numbers
Problem 6 558 views 
Dividing Complex Numbers
Problem 7 269 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete