4.1 Exponential and Logarithmic Functions
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1 . Exponential and Logarithmic Functions Joseph Heavner Honors Complex Analysis Continued) Chapter July, 05 3.) Find the derivative of f ) e i e i. d d e i e i) d d ei ) d d e i ) e i d d i) e i d d i) iei + ie i 7.) Write arg e i+)) in terms of x and y. By 5) we have the argument is simply Ii + )), so we simply expand this to get the desired result. Ii + )) Iix + iy) + x iy))) Iix) x But, we must recall periodicity, which provides the final solution x + nπ, n 0, ±, ±,... 5.) Find the image of the line y under the exponential mapping. Horiontal lines map to rays, in particular the line y c Argw) c, thus the solution is simply Argw) 3.) Find all complex values of ln + i). The complex natural logarithm is simply lnr) + iθ, and we have that r 8 e 3/ and θ 3π so the logarithm is ) 3π ln 3/ ) + i + nπ 3 8n + 3)π ln) + i, n 0, ±, ±,... + nπ, 7.) Find the principal value of Ln6 6i) in a + bi form. Here we restrict π < θ π and have r 7 and θ π/, so Ln6 6i) ln 7) iπ ln7) π i 35.) Find all C satisfying e ie 3. This implies that ln ie 3 ), which means + ln ie 3 ). We simplify using Definition..: + lne 3 ) + i π ) ) n )π + nπ + 3 lne) + i + n )π i, n 0, ±, ±,...
2 3.) Find the image of the circle under the mapping w Ln). Circles map to vertical lines under this mapping, in particular this maps to the vertical line segment u ln), π < v π 7.) Use ) to prove that e /e e. This and similar properties are simple enough to prove. e /e ex cosy ) + i siny )) e x cosy ) + i siny )) But we know division of complex numbers amounts to dividing moduli and subtracting arguments, so we have And this gives us the desired e. e /e e x x [cosy y ) + i siny y )]
3 . Complex Powers 3.) Find all values of + i) i. and θ π/ + nπ, thus So, we have ln + i) ln ) + i π/ + nπ) 8n + )π ln) + i, n 0, ±, ±,... + i) i e i) ln+i) e i) 8n+)π ln)+ i) e ln))/+8n+)π/+i[8n+)π/ ln))/] Noting that e ln)/ yields the simplified a misnomer here) solution + i) i 8n + )π 8n + )π exp + i ln) )), n 0, ±, ±,... 7.) Find the principal value of ) 3i. This is less messy than the last problem. and Arg) π, so Ln ) ln) + iπ iπ. Thus, ) 3i e 3iLn ) e 3π 5.) Find the derivative of 3/ on the domain > 0, π < arg) < π at + i. Since + i is in the function s usual domain we have that f ) +i 3 / +i 3 + i)/ If one were so inclined, one could also find that + i e πi 8, so 3/ ) 3 eπi/8 3
4 .3 Trigonometric and Hyperbolic Functions 7.) Express sec π i ) in terms of a + bi. π ) sec i cos π i ) cosπ/) cosh ) i sinπ/) sinh ) i sinh) icsch).) Find all C such that sin cos. Using the definition of the trigonometric functions in terms of the exponential and skipping a few steps) e i e i ei + e i i e i ie i + i e i ) i Taking the square root of i yields e iπ/ and e 5iπ/, so we solve two equations, setting both of those equal to e i. To save some time, the work will only be show for the first case. e i e iπ/ π ) i ln) + i + πk + 8k)π For the second case we have a similar solution: Thus,, k 0, ±, ±,..., 7π, π, 9π,, 3π, 5π, 3π, n + )π, n 0, ±, ±,... 5.) Verify that cos cos C. cos cosx) cosh y) i sinx) sinh y) cosx) coshy) + i sinx) sinhy) cosx) coshy) i sinx) sinhy) cos 9.) Find d d tan ) ). d tan ) d ) tan ) + tan ) sec ) ) + tan ) ) sec + tan )
5 3.) Express cosh + π 6 i) in a + bi form. cosh + π i 6 i) cos + π )) 6 i π ) 6 + i cos cos π ) cosh) i sin π ) sinh) cosh) + i sinh) 7.) Find all C such that sinh cosh. By definition Therefore, no solutions exist. e e e + e e e e ) e e + e ) 3.) Verify that sinh sinh x + sin y. sinh i sini) sin y + ix) sin y) + sinh x) sin y) + sinh x sin y) + sinh x) 35.) Find the derivative of tanhi ). tanhi )) sech i ) i ) sech i ) i isech i ) 5
6 . Inverse Trigonometric and Hyperbolic Functions 7.) Find all values of sinh i. sinh i) ln i + + ) /) ln) + i π n + )π + nπ) i, n 0, ±, ±,... 5.) Use the branch re iθ/, π < θ < 0 and the principal branch of ln to a) find the value of cosh at i, and b) find the derivative of cosh at that point. a) cosh i) ln i + ) /) ln i + ) /) ln i + e iπ/) ln ) + )i ) ln + π i b) cosh )) i ) / i ) / e iπ i 9.) Use implicit differentiation to derive formula 8) for the derivative of the of the inverse cosine. Note that w cos ) cosw), so we implicitly differentiate to get sin dw d dw d sin w But, sin w + cos w sin w +, so sin w. Thus, dw d sin w 6
7 Special Topic: The Lambert W Function While we are on the topic of the exponential function and its inverse, the logarithmic function, it is worth briefly discussing the famous Lambert W. This function can be defined by the following W)e W), C As with most complex functions, this is multi-valued. However, we can restrict the function to be single-valued. While this function generally comes in play in combinatorics or applied settings, we shall look at a single example from the simple setting of exponential equations. This is a particularly significant equation, one that is not only common but also instructive. Example Solve the equation x x. We begin with logarithms Now, we invoke the W function. x x x ln x ln e ln x ln x ln e ln x ln x ln ln x Wln ) Wln ) x e It is worth looking into this further, if only just for entertainment value. provides a good image of the function s Riemann surface as well as some examples, cbeldp is a good technical exposition, and finally provides an extended, somewhat technical introduction as well as some history. 7
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