Solutions for assignment 2

# Solutions for assignment 2 - Solutions for assignment 2 1...

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Section 2.2 Question 9: (a) (z-5i)^2 is a polynomial, hence continuous everywhere, and thus on e can simply substitute in the limit to obtain ((2+3i)^2 - 5i)^2 = -8i. (c) This is an indeterminate form; one can either use L'hopital's rule (which is valid for complex functions, although we will not prove this until much later), or factorize the numerator as z^2 + 9 = (z-3i)(z+3i) and simplify the limit to z+3i. Now the function is continuous and we can simply substitute to obtain 6i. (e) The function in the limit can be simplified to 2z_0 + /\z, which (being a polynomial in /\z) converges to 2z_0 as /\z tends to zero. Alternatively, one can note that this is precisely the derivative of z^2 at z_0, which is equal to 2z_0 since the differentiation of polynomials proceeds in complex analysis in exactly the same manner as in real analysis. Question 15: The limit along the real approach y=0 is 2i, and the limit along the imaginary approach x=0 is 2i+1, which are not equal, so the full limit does not exist. Section 2.3 Question 4: (a) The expression in the limit of Definition 4 simplifies to /\x / /\ z, where /\z = /\x + i /\y is the Cartesian form of /\z. Taking the limits along the real and imaginary approaches shows that this limit does not exis t. (b) is proven in the same way as (a).

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## This note was uploaded on 09/19/2011 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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Solutions for assignment 2 - Solutions for assignment 2 1...

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