For this exercise, || stands for complex magnitude rather than order of a group element. Let...
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For this exercise, || stands for complex magnitude rather than order of a group element. Let z, w E C. Show that |zw| = |z||w|. Show that |zn| = |2|" for all n E Z (this is one of those induction-for-positive-values- of-n things.) Conclude that if |z| # 1, then z has infinite order in Cx. For this exercise, || stands for complex magnitude rather than order of a group element. Let z, w E C. Show that |zw| = |z||w|. Show that |zn| = |2|" for all n E Z (this is one of those induction-for-positive-values- of-n things.) Conclude that if |z| # 1, then z has infinite order in Cx.
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To show that w w we need to show that w w for any complex number w By definition w z in CX zn 1 for ... View the full answer
Related Book For
Income Tax Fundamentals 2013
ISBN: 9781285586618
31st Edition
Authors: Gerald E. Whittenburg, Martha Altus Buller, Steven L Gill
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