Math 132 midterm

Math 132 midterm - Solutions to first midterm, 132/1 Winter...

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Problem 1 (10 points): Prove that cos(2z) = cos^2(z) - sin^2(z) for all complex numbers z. Answer: Write the left-hand side as [exp(2iz) + exp(-2iz)] / 2 and the left-hand side as ([exp(iz) + exp(-iz)] / 2)^2 - ([exp(iz) - exp(-iz)] / 2i)^2. If you expand both sides and simplify, the two sides will agree, which proves the claim. Some common errors: Expanding (a+b)^2 as a^2 + b^2, or (a-b)^2 as a^2 - b^2. Forgetting about various i, especially in the denominator in the definition of sin(z). Interchanging the definitions of sin(z) and cos(z). Other comments: Some of you showed that the identity followed from the complex cosine addition law cos(z+w) = cos(z) cos(w) - sin(z) sin(w) This is true, but it begs the question of why the complex cosine addition law is true. To be true to the spirit of the question, you would then have to prove the complex cosine addition law, perhaps by breaking up into exponentials. Some of you tried to expand cos(z) = cos(x+iy) using the cosine addition law. This is not exactly an error, but it is not really progress either. It also begs the question as before. Some of you expanded the left-hand side only, and tried to turn it into the right-hand side. This will work if you know what you're doing, but in this case it is far easier to start with the right-hand side and expand that. (Actually, for these problems it is usually best to expand out both sides and try to get them to meet somewhere in the middle). Problem 2(a) (5 points): Find an entire function f whose real part is equal to x^2 + xy - y^2. Answer: Write f = u+iv. We are given that u(x+iy) = x^2 + xy - y^2. Since f is entire, we have the Cauchy- Riemann equations partial u / partial x = partial v / partial y, partial u / partial y = - partial v / partial x which in our case simplifies to partial v / partial y = 2x - y, partial v / partial x = 2y - x. Integrating the first equation one gets Solutions to first midterm, 132/1 Winter 2000 1 of 7 9/19/2011 3:46 PM
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v(x+iy) = 2xy - y^2/2 + c(x) where c(x) is a function that depends only on x (it must be constant in the y direction). Inserting this into the second equation we get 2y + c'(x) = 2y - x which solves to c(x) = -x^2/2 + C where C is now a genuine constant (something that does not depend on either x or y). So v(x+iy) = 2xy - x^2/2 - y^2/2 + C and so f(x+iy) = u(x+iy) + i v(x+iy) = x^2 + xy - y^2 + i(2xy - x^2/2 - y^2/2 + C). Since we only need to find one such function f, we can set C to be any value we want, say 0. So one possible answer is f(x+iy) = u(x+iy) + i v(x+iy) = x^2 + xy - y^2 + i(2xy - x^2/2 - y^2/2). We can then verify that f is analytic by rechecking the Cauchy-Riemann equations. One could also try to simplify the right-hand side in terms of z; it turns out that we can re-arrange the above as f(z) = (1 - i/2) z^2 which is clearly entire (it's a polynomial in z). Some common errors:
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Math 132 midterm - Solutions to first midterm, 132/1 Winter...

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