Problem 5★. This problem walks through a straightforward proof of Euler's formula from Problem 6. The proof pro- ceeds as follows: Trusting that e¹x is a complex number, we can (just like any other complex number) represent it in polar coordinates. Let r, 0 ER with r > 0. Our goal then is to simply find these unknown values r and 0. We expect them to depend on whatever x is, and thus we want to think of two real-valued functions 0(x) and r(x). So we have eix * = r(x) (cos 0(x) + i sin 0(x)) (1) We can differentiate² both sides with respect to x, using the product rule for differentiation on the right-hand side: d d dx dx [] = [(x r(x) (cos(x)+isin(x) (2) dr d ⇒ieix [cos 0(x)+isin 0(x)] + r(x)· dx [cos 0(x)+isin 0 (x)] (3) dx ⇒ieix dr de [cos 0(x)+isin 0(x)] + r(x) [-sin 0 (x) + i cos 0 (x)] (4) dx dx where dry and do are some mathematical expressions to be determined. (a) Use Eq. (1) to rewrite the left side of Eq. (4). Then, rearrange each side of the equation, grouping the real and imaginary components together. (b) We know the real part of the left-hand side must equal the real part of the right-hand side, and the imaginary part of the left-hand side must equal the imaginary part of the right-hand side. Use these facts to build a 2-by-2 system of equations. (c) Examine this system of equations and conclude that dr = 0 and do = 1. This means that r(x) is a constant (let's call it R) and 0(x) is a first-degree polynomial, 0(x) = x+C for some unknown constant C. Substitute these into Eq. (1) in place of 0(x) and r(x). (d) Considering the case when x = 0, we can use the fact that e°i = e° = 1 to determine what R and C must be (again by equating the real components of each side, and then the imaginary components of each side). By choosing the correct values of R and C, we can simplify to yield Euler's formula, and the proof is done!