# Video: Solving Equations Involving Complex Numbers in Polar Form

Given that ππ^(2ππ) + ππ^(β2ππ) = cos (2π) β 5π sin (2π), where π β β and π β β, find π and π.

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### Video Transcript

Given that ππ to the two ππ plus ππ to the negative two ππ is equal to cos two π minus five π sin two π, where π and π are real numbers, find π and π.

In order to answer this question, we need to ensure that the complex numbers on both sides of the equation are in the same form. Currently, the left-hand side is in exponential form and the right-hand side in polar form. Letβs begin by rewriting the complex numbers ππ to the two ππ and ππ to the negative two ππ in polar form.

We know that ππ to the ππ is equal to π multiplied by cos π plus π sin π, where π is the modulus and π is the argument of the complex number. The modulus of our first complex number is equal to π and the argument is two π. Therefore, ππ to the two ππ is equal to π multiplied by cos two π plus π sin two π. In the same way, ππ to the negative two ππ is equal to π multiplied by cos of negative two π plus π sin of negative two π.

We know that cos π is an even function, and sin π is an odd function. Therefore, cos of negative two π is equal to cos two π, and sin of negative two π is equal to negative sin of two π. This means that we can rewrite the equation as π multiplied by cos two π plus π sin two π plus π multiplied by cos two π minus π sin two π.

Next, we can distribute π and π over their parentheses, as shown. We can then collect like terms and factor out cos two π and sin two π. This gives us π plus π multiplied by cos two π plus π minus π multiplied by π sin two π. Comparing this to our original equation, we know this is equal to cos two π minus five π sin two π. Equating the coefficients of cos two π gives us π plus π is equal to one, and equating the coefficients of sin two π gives us π minus π is equal to negative five.

We have a pair of simultaneous equations. We can solve by elimination by adding the two equations. This gives us two π is equal to negative four. Dividing both sides of this by two gives us π is equal to negative two. We can then substitute this value back into equation one such that negative two plus π is equal to one. Adding two to both sides of this equation gives us π is equal to three.

We now have the values of the two constants. π is equal to negative two and π is equal to three.