# Video: Dividing Complex Numbers in Algebraic Form and Expressing Their Quotient in Exponential Form

Given that π§β = 1/2 β (β(3)/2)π and π§β = 2β(3) + 2π, find π§β/π§β, giving your answer in exponential form.

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### Video Transcript

Given that π§ one is equal to one-half minus root three over two π and π§ two is equal to two root three plus two π, find π§ one divided by π§ two, giving your answer in exponential form.

Weβll begin by expressing π§ one divided by π§ two as one-half minus root three over two π all over two root three plus two π. And then, we recall that to evaluate the quotient of two complex numbers in rectangular form, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of a complex number is found by changing the sign between the two terms. For π plus ππ, the conjugate is π minus ππ.

This means that the conjugate of two root three plus two π is two root three minus two π. So we multiply both the numerator and the denominator by two root three minus two π. Letβs do this in two parts. Weβll expand these brackets as normal. We begin by multiplying the first term in each bracket. One-half multiplied by two root three is simply root three. We then multiply the outer terms. One-half multiplied by negative two π is simply negative π. Then multiplying the inner terms, negative root three over two π multiplied by two root three is root three squared π. And of course, root three squared is three. So this becomes negative three π. And then, we multiply the last terms. Negative root three over two π multiplied by negative two π is positive root three π squared. And since π is the square root of negative one, π squared is negative one. And this becomes negative root three.

So our brackets expand to root three minus π minus three π minus root three. Root three minus root three is zero. And we are left with negative four π. Weβll now repeat this process to multiply two root three plus two π by two root three minus two π. Multiplying the first two terms, we get four root three squared and since root three squared is three this becomes four multiplied by three which is 12. Two root three multiplied by negative two π is negative four root three π. Two π multiplied by two root three is four root three π. And of course, two π multiplied by negative two π is negative four π squared. And since π squared is negative one, this becomes plus four. The negative four root three π and positive four root three π cancel each other out and weβre left with 16.

So the quotient of π§ one and π§ two is negative four π over 16, which we can further simplify to negative π over four. Now that we have π§ one divided by π§ two in rectangular form, letβs find a way to write this in exponential form. We write the complex number as π π to the ππ, where π is the modulus or sometimes called the magnitude of our complex number. Itβs found by finding the square root of π squared plus π squared. And π is arctan of π over π and that must be in radians.

If we compare our complex number to the general form, we can see that π is zero. π is the coefficient of π and in this case itβs negative one-quarter. So the modulus π is the square root of zero squared plus negative one-quarter squared which is one-quarter. If we then substitute what we know about our complex number into the formula for π, we get arctan of negative one quarter over zero. Now, negative one-quarter divided by zero is undefined. Now, here we need to recall the general result that tan of π over two or tan of negative π over two is undefined. So which one do we choose?

Well, to decide, we recall the Argand diagram. We can see the horizontal axis represents the real component and the vertical axis represents the imaginary component of our complex number. Our complex number is zero plus negative one-quarter π. Thatβs here. So since we measure from the horizontal axis in a counterclockwise direction, we could say that this angle is negative π over two. Or alternatively, we could say that itβs three π over two. Negative π over two is the principal argument. Here, weβre going to choose three π over two.

So we can say that our complex number π§ one over π§ two in exponential form is a quarter π to the power of three π by two π.