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Unformatted text preview: given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back√ rectangular form – certainly more work than is required to √ to multiply out zw = (2 3 + 2i)(−1 + i 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of complex numbers. Consider how we computed w5 above and compare that √ using the Binomial Theorem, Theorem 9.4, to accomplish the same feat by to expanding (−1 + i 3)5 . Division is tricky in the best of times, and we saved ourselves a lot of z time and effort using Theorem 11.16 to find and simplify w using their polar forms as opposed to starting with √ 2 3+2i √, −1+i 3 rationalizing the denominator, and so forth. There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11.16. Take the product rule, for instance. If z = |z |cis(α) and w = |w|cis(β ), the formula zw = |z ||w|cis(α + β ) can be viewed geometrically as a two st...
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