Video: Solving Equations Involving Complex Numbers in Polar Form

Given that π‘Žπ‘’^(2π‘–πœƒ) + 𝑏𝑒^(βˆ’2π‘–πœƒ) = cos (2πœƒ) βˆ’ 5𝑖 sin (2πœƒ), where π‘Ž ∈ ℝ and 𝑏 ∈ ℝ, find π‘Ž and 𝑏.

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Video Transcript

Given that π‘Žπ‘’ to the two π‘–πœƒ plus 𝑏𝑒 to the negative two π‘–πœƒ is equal to cos two πœƒ minus five 𝑖 sin two πœƒ, where π‘Ž and 𝑏 are real numbers, find π‘Ž 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 π‘Žπ‘’ to the two π‘–πœƒ and 𝑏𝑒 to the negative two π‘–πœƒ in polar form.

We know that π‘Ÿπ‘’ to the π‘–πœƒ is equal to π‘Ÿ multiplied by cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument of the complex number. The modulus of our first complex number is equal to π‘Ž and the argument is two πœƒ. Therefore, π‘Žπ‘’ to the two π‘–πœƒ is equal to π‘Ž multiplied by cos two πœƒ plus 𝑖 sin two πœƒ. In the same way, 𝑏𝑒 to the negative two π‘–πœƒ is equal to 𝑏 multiplied by cos of negative two πœƒ plus 𝑖 sin of negative two πœƒ.

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

Next, we can distribute π‘Ž and 𝑏 over their parentheses, as shown. We can then collect like terms and factor out cos two πœƒ and sin two πœƒ. This gives us π‘Ž plus 𝑏 multiplied by cos two πœƒ plus π‘Ž minus 𝑏 multiplied by 𝑖 sin two πœƒ. Comparing this to our original equation, we know this is equal to cos two πœƒ minus five 𝑖 sin two πœƒ. Equating the coefficients of cos two πœƒ gives us π‘Ž plus 𝑏 is equal to one, and equating the coefficients of sin two πœƒ gives us π‘Ž 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 π‘Ž is equal to negative four. Dividing both sides of this by two gives us π‘Ž 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. π‘Ž is equal to negative two and 𝑏 is equal to three.

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