# sol8 - MAE294B/SIO203B Methods in Applied Mechanics Winter...

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MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution VIII 1 Find the points where the radicals vanish: z 2 + 1 = 0 gives ± i; 2 - z 2 + 1 = 0 gives ± 3. There are hence 4 branch points and many ways of joining them. The default Matlab branch cut takes z 2 + 1 = - u 2 with u > 0, i.e. the imaginary axis with | y | > 1 and z 2 + 1 = ( 2 + v 2 ) 2 with v > 0, i.e. the real axis with | x | > 3. There are potentially subtle issues if a branch of z 2 + 1 is chosen that is everywhere negative; ignore these. Figure 1 shows the real and imaginary parts of two different branches computed by Matlab. Figure 1: (a) and (b): real and imaginary parts of sqrt(2 - sqrt(z.^2+1)) . (c) and (d): real and imaginary parts of sqrt(2 - i sqrt(-z.^2-1)) . 2 We have to choose a branch of ( z 2 - 1 ) 1 / 2 . The obvious one has a cut along the real axis between - 1 and 1, and is positive and real on the real axis with x > 1. Consider the integral Z C d z z 2 - 1 around a contour C surrouding the cut. Shrinking the contour onto the cut gives Z 1

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