Question: For this exercise, || stands for complex magnitude rather than order of a group element. Let z, w E C. Show that |zw| =

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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