can-soln4.pdf - MATH 20142 Complex Analysis Solution Sheet...

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MATH 20142 Complex Analysis Solution Sheet 4 1. (a), (b) Since cos( - z ) = cos z, sin( - z ) = - sin z , we have e iz = cos z + i sin z, e - iz = cos z - i sin z, and (a), (b) follow from adding and subtracting these two identities and subsequent dividing by two. (c) From (a) and (b) it follows that cos 2 z = 1 4 ( e 2 iz + 2 + e - 2 iz ) , sin 2 z = - 1 4 ( e 2 iz - 2 + e - 2 iz ) , whence cos 2 z + sin 2 z = 1. (d) By (b), sin( z + w ) = 1 2 i ( e i ( z + w ) - e - i ( z + w ) ) . ( * ) On the other hand, sin z cos w = 1 2 i ( e iz - e - iz ) · 1 2 ( e iw + e - iw ) = 1 4 i ( e iz e iw + e iz e - iw - e - iz e iw - e - iz e - iw ) and cos z sin w = 1 2 i ( e iw - e - iw ) · 1 2 ( e iz + e - iz ) = 1 4 i ( e iw e iz + e iw e - iz - e - iw e iz - e - iw e - iz ) Summing these two identities yields (*). (e) Similar to (d). 2. It has been shown in the lectures that | sin z | 2 = sin 2 x + sinh 2 y. Now, if β = 0, then x = 0, and we have | sin z | = | sinh y | , which is unbounded. If β ̸ = 0, then y = kx , where k = - α/β ; hence | sin z | 2 = sin 2 x + sinh 2 ( kx ) sinh 2 ( kx ) . We see that if k ̸ = 0, then | sin z | is unbounded, as | sinh kx | is unbounded; if k = 0 (whence α = 0), then | sin z | = | sin x | ≤ 1.
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  • Spring '14
  • Sidorov
  • Math, Equations, Cos, Elementary algebra, Hyperbolic function, eiz

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