## mginaction 3 years ago Unit 2, Exponential Response, Sinusoidal Input PDF, page 1 at the bottom: I am confused by how they got from 3+16i = sqrt(265)e^i@ (@ = phi) to the solution 9/sqrt(265)cos(2t-@). I feel fairly comfortable with the complex numbers, Eulers Formula etc, and I also understand the Exponential Response Formula. It's just that step that I am confused about. I used complex conjugate to multiply out 9/(3+16i) then put into exponential form (9/sqrt(265))(cos@ - isin@), which makes the solution the real part of ((9/sqrt(265))(cos@ - isin@)(cos2t + isin2t). I just dont understand how that ends up being (

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1. mginaction

I dont understand how that ends up being 9/sqrt(265) * cos(2t - @). and yes I undertand cos(t) + sin(t) = cost(t - @). I am taking ODE at UVA Wise, but my professor isn't teaching complex replacement. Any help would be seriously appreciated. I really want to learn this method.

2. wrwrwr

The complex solution is $$z_p = \frac{9 e^{2it}}{3 + 16i}$$, and what we want is its real part in the so-called amplitude-phase form. -- Start by transforming the denominator to the polar form: $$3 + 16i = \sqrt{3^2 + 16^2} e^{i \, atan2(16, 3)} = \sqrt{265} e^{i \phi}$$, where $$\phi = tan^{-1}(16/3)$$; -- (Don't yet use Euler's definition); -- Your solution is now: $$z_p = \frac{9}{\sqrt{265}} \frac{e^{2it}}{e^{i \phi}}$$; -- Now use a simple exponents law ($$x^n / x^m = x^{n-m}$$) to get $$z_p = \frac{9}{\sqrt{265}} e^{i (2t - \phi)}$$; -- Use Euler's formula: $$z_p = \frac{9}{\sqrt{265}} (cos(2t - \phi) + i \, sin(2t - \phi))$$; -- And drop the imaginary part: $$x_p = Re(z_p) = \frac{9}{\sqrt{265}} cos(2t - \phi)$$. You could also expand the numerator using Euler's, expand the fraction by denominator's conjugate, multiply brackets (leaving the conjugate in Cartesian form and using $$i^2 = -1$$), get the real part and use the trigonometric formula ($$a \, cos(\theta) + b \, sin(\theta) = A \, cos(\theta - \phi)$$ with $$a + ib = A \, e^{i \phi}$$).