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November 3, 2010 Lecturer Dmitri Zaitsev Michaelmas Term 2010 Course 2325 2010 Complex Analysis I Sheet 2 Due: at the end of the lecture Exercise 1 Consider the sequence of holomorphic functions fn (z) = z + n1 . (i) Is the sequence (fn ) converging uniformly on C? (ii) Is the sequence of squares (fn2 ) converging uniformly on C? Justisfy your answer. Solution (i) The sequence (fn ) has pointwise limit f (z) = z. Then the convergence is uniform on C if and only if sup |fn (z) − f (z)| → 0, n → ∞. z∈C In our case supz∈C |fn (z) − f (z)| = 1 n, hence the convergence is uniform. (ii) The sequence of squares (fn2 ) converges pointwise to the square f 2 . However, this time sup |fn (z)2 − f (z)2 | = sup | z∈C z∈C 2z 1 + 2 | = +∞, n n hence the left-hand side does not converge to 0 and therefore the convergence is not uniform. Exercise 2 Use the Cauchy-Riemann equations to decide which of the following functions are holomorphic: Imz, −i|z|2 , z̄ 2 , ez+i , ez̄ . Solution We use the complex form of the Cauchy-Riemann equation: ∂z̄ f = 0, where ∂z̄ = 12 (∂x + i∂y ). Then we obtain 2∂z̄ (Imz) = (∂x + i∂y )y = i 6= 0, 2∂z̄ (−i|z 2 |) = (∂x + i∂y )(−ix2 − iy 2 ) = −2ix + 2y 6= 0, 2∂z̄ (z̄ 2 ) = (∂x + i∂y )(x2 − 2ixy − y 2 ) = 2x − 2iy + 2x − 2iy 6= 0 2∂z̄ (ez+i ) = (∂x + i∂y )(ex (cos(y + 1) + isin(y + 1))) = = ex (cos(y + 1) + isin(y + 1)) + iex (−sin(y + 1) + icos(y + 1) = 0, 2∂z̄ (ez̄ ) = (∂x + i∂y )(ex (cosy − isiny)) = = ex (cosy − isiny) + iex (−siny − icosy) 6= 0, thus only ez+i is holomorphic.