Lesson: Division of Complex Numbers

Question 2

1 point


One of our first uses of complex conjugation will be to help us divide complex numbers. Remember that division is simply an inverse operation for multiplication. That is, we say that a/b = c if and only if a = bc.

In the complex numbers, we want to notice the following. If z1 = z2z3, then (z2-conjugate)z1 will equal (z2-conjugate)(z2z3), which means that (z2-conjugate)z1 will equal (x2-squared + y2-squared)z3. And so this means that ((z2-conjugate)z1)/(x2-squared + y2-squared) = z3. And so we can write z3 = z1/z2 = (z1(z2-conjugate))/(x2-squared + y2-squared).

We expand this formula even further to get the following definition. The quotient of two complex numbers z1 = x1 + y1i and z2 = x2 + y2i is z1/z2, which equals (z1(z2-conjugate))/(z2(z2-conjugate)), which equals ((x1x2 + y1y2)/(x2-squared + y2-squared)) + ((y1x2 – x1y2)/(x2-squared + y2-squared))i.

I should like to note that I don’t think anyone ever bothers to memorize this formula. We simply go through the process of multiplying the top and bottom by the conjugate each time.

So for example, if we wanted to look at the quotient (3 – 2i)/(1 + 4i), then we would multiply the top and bottom by (1 + 4i)-conjugate, which is 1 – 4i. Multiplying this out on the top, we’ll get 3(1) + 3(-4i) + (-2i)(1) + (-2i)(-4i), and on the bottom, we know we’re going to get that x2-squared + y2-squared, so we don’t really have to multiply it out. We know it will be 1-squared + 4-squared. So on our top, we see that our real part becomes a (3 – 8), while our imaginary part becomes the (-12 – 2)i, and our bottom is a (1 + 16), and so we see that the quotient is –(5/17) – (14/17)i.

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