# Question Video: Simplifying Complex Number Expressions Using Conjugates and Converting Them to Polar Form Mathematics

Simplify (6 β 6π)/β2π, giving your answer in both algebraic and trigonometric form.

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### Video Transcript

Simplify six minus six π over negative two π, giving your answer in both algebraic and trigonometric form.

To divide two complex numbers in rectangular form, thatβs the general form π§ equals π plus ππ, we first need to find the complex conjugate of the denominator. For a complex number π plus ππ, its complex conjugate is π minus ππ. The denominator of our fraction is negative two π. We can think of that as zero minus two π, and this means that the complex conjugate of the denominator is zero plus two π is actually just two π. To evaluate this fraction then, we multiply both the numerator and the denominator by two π. Six multiplied by two π is 12π, and negative six π multiplied by two π is negative 12 π squared. Negative two π multiplied by two π is negative four π squared. Remember though, π squared is just negative one, so our fraction simplifies to 12 π plus 12 all over four.

We can divide through by this four and we get three π plus three. Notice Iβve actually written it as three plus three π to match the general algebraic form for a complex number. In algebraic or rectangular form then, the complex number is π§ is equal to three plus three π. And if we compare this to the general form, we can see that π, the real component, has a value of three and π, the imaginary component, also has a value of three. We need to find a way to write this complex number in trigonometric or polar form; thatβs π§ is equal to π cos π plus π sin π. π is the modulus of a complex number π§, and π is the argument in polar form or trigonometric form. π can be in degrees or radians. In exponential form though, it does need to be in radians, so we need to find a way to represent the real and component parts of our number π§ in terms of π and π.

In fact, we can use this formulae to help us: the modulus π is equal to the square root of π squared plus π squared; and to find π, we use tan π is equal to π over π. Letβs substitute what we know about our complex number into these formulae. π is three and π is three. So the modulus π is the square root three squared plus three squared, which is the square root of 18. This simplifies to three root two. Tan π is three divided by three or one. We can solve this by finding the arctangent of one, which is π over four. And then we can substitute all values for π and π into the general trigonometric form of a complex number. And doing so, we get that π§ is equal to three root two cos of π over four plus π sin of π over four. And remember, in our algebraic or rectangular form, we said that π§ was equal to three plus three π.