hw2_soln - Complex Analysis Fall 2007 Homework 2: Solutions...

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Complex Analysis Fall 2007 Homework 2: Solutions 1.3.6. (a) We have log | - i | = log 1 = 0 and arg( - i ) ∈ {- π/ 2 + 2 | n Z } . Hence, the values of log( - i ) are i ± - π 2 + 2 for n Z . (b) We have log | 1 + i | = log 2 = (1 / 2)log 2 and arg(1 + i ) ∈ { π/ 4 + 2 | n Z } . Hence, the values of log(1 + i ) are 1 2 log 2 + i π 4 + 2 · . 1.3.8. (a) Since i is not a rational number, we know that ( - 1) i has infinitely many values. To compute them we first compute the values of log( - 1). Since log | - 1 | = log 1 = 0 and arg( - 1) ∈ { π +2 | n Z } the values of log( - 1) are i ( π +2 ) for n Z . Hence, the values of ( - 1) i are e i log( - 1) = e i 2 ( π +2 ) = e - ( π +2 ) for n Z . (b) Again, i is not rational so we expect infinitely many values for 2 i . As above, we have log | 2 | = log 2 and arg 2 ∈ { 2 | n Z } so that log 2 = log 2 + 2 nπi and 2 i = e i log 2 = e - 2 + i log 2 for n Z . 1.3.12. We begin by observing that for any α C we have (by definition) tan α = sin α cos α = - i e - e - e + e - . If we multiply the numerator and denominator by e this becomes tan α = - i ( e ) 2 - 1 ( e ) 2 + 1 . Given
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This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.

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hw2_soln - Complex Analysis Fall 2007 Homework 2: Solutions...

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