# Video: Solving Equations Involving Complex Numbers in Polar Form

Given that 𝑎𝑒^(2𝑖𝜃) + 𝑏𝑒^(−2𝑖𝜃) = cos (2𝜃) − 5𝑖 sin (2𝜃), where 𝑎 ∈ ℝ and 𝑏 ∈ ℝ, find 𝑎 and 𝑏.

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### Video Transcript

Given that 𝑎𝑒 to the two 𝑖𝜃 plus 𝑏𝑒 to the negative two 𝑖𝜃 is equal to cos two 𝜃 minus five 𝑖 sin two 𝜃, where 𝑎 and 𝑏 are real numbers, find 𝑎 and 𝑏.

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. Let’s begin by rewriting the complex numbers 𝑎𝑒 to the two 𝑖𝜃 and 𝑏𝑒 to the negative two 𝑖𝜃 in polar form.

We know that 𝑟𝑒 to the 𝑖𝜃 is equal to 𝑟 multiplied by cos 𝜃 plus 𝑖 sin 𝜃, where 𝑟 is the modulus and 𝜃 is the argument of the complex number. The modulus of our first complex number is equal to 𝑎 and the argument is two 𝜃. Therefore, 𝑎𝑒 to the two 𝑖𝜃 is equal to 𝑎 multiplied by cos two 𝜃 plus 𝑖 sin two 𝜃. In the same way, 𝑏𝑒 to the negative two 𝑖𝜃 is equal to 𝑏 multiplied by cos of negative two 𝜃 plus 𝑖 sin of negative two 𝜃.

We know that cos 𝜃 is an even function, and sin 𝜃 is an odd function. Therefore, cos of negative two 𝜃 is equal to cos two 𝜃, and sin of negative two 𝜃 is equal to negative sin of two 𝜃. This means that we can rewrite the equation as 𝑎 multiplied by cos two 𝜃 plus 𝑖 sin two 𝜃 plus 𝑏 multiplied by cos two 𝜃 minus 𝑖 sin two 𝜃.

Next, we can distribute 𝑎 and 𝑏 over their parentheses, as shown. We can then collect like terms and factor out cos two 𝜃 and sin two 𝜃. This gives us 𝑎 plus 𝑏 multiplied by cos two 𝜃 plus 𝑎 minus 𝑏 multiplied by 𝑖 sin two 𝜃. Comparing this to our original equation, we know this is equal to cos two 𝜃 minus five 𝑖 sin two 𝜃. Equating the coefficients of cos two 𝜃 gives us 𝑎 plus 𝑏 is equal to one, and equating the coefficients of sin two 𝜃 gives us 𝑎 minus 𝑏 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 𝑎 is equal to negative four. Dividing both sides of this by two gives us 𝑎 is equal to negative two. We can then substitute this value back into equation one such that negative two plus 𝑏 is equal to one. Adding two to both sides of this equation gives us 𝑏 is equal to three.

We now have the values of the two constants. 𝑎 is equal to negative two and 𝑏 is equal to three.