MATH 113 Midterm: MATH 113 Harvard 113 Fall 01113Solution3

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15 Feb 2019
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Since eiy = cos y + i sin y, one sees that for any real y0, eiy maps the interval [y0, y0 + 2 ) bijectively onto the unit circle. Now ex maps the real line bijectively onto the set of positive real numbers. Now if yi is a sequence of real numbers approach- ing y0 + 2 from below, then clearly limi logy0(eiyi) = y0 + 2 . We have shown that the function logy0 does not preserve limits of sequences, so it is not continuous. We see that logy0 z1z2 = log|z1z2| + i , where is some real number such that ei = z1z2. In a similar manner we can write logy0 z1 = log|z1| + i 1 and logy0 z2 = log|z2| + i 2. |z2| , we see that ei( 1+ 2) = z1z2. Hence and 1 + 2 di er by a multiple of 2 .