ANSWERS 5975 e. Note that z^2 - z + 2 = 0 when z = i±;Y!. Also, Arg ( l±;Y!) =
±Arc tan ( J7) , respectively. For a = Arc tan ( ./7) , the function w =
f ( z) = Jog" ( z^2 - z + 2) is differentiable for z = r ei^8 f. i-;Y!, where°' < 0 < 0!+2n. For a= -Arctan(v'7) the function w = f(z) =
log" (z^2 - z + 2) is differentiable for z = rei^8 f.^1 +;n, where °' < 0 <
°' + 2n. Furthermore, in each case ~~ = f' (z) = }:~! 2 for z in the
respective domains.7a. In (x^2 + y^2 ) = 2Re (Log (z)), and Log is analytic for Re (z) > O.
9. According to equation (5-20), f (z) = Jog 6 ,. (z + 4) has f (-5) = log 6 " (-1) =
lnl- 11+i(7n)=7ni.lla. The function f (z) =lo~ (z + 2) does the job. Explain why.
llc. The function f (z) = log_i (z + 2) works. Explain why.
- There a.re many possibilities, such as z 1 =1, Zz = - 1. Explain.
1 5. Any branch of the logarithm is defined as an inverse of the exponential.
Since there is no value z for which exp (z) = 0, there can be no branch of
the logarithm that is defined at 0.Sect io n 5.3. Complex Exponents: page 175la. cos (Jn4) + isin (Jn4).
le. cos 1+isin1.- Note that O·log(z) = {O -(: ( E Jog(z)}. This collapses to the single element
zero. Thus, for z f. 0, z^0 = exp (0 ·log z) = exp (0) = 1.
s. 2z..- 1 -2zn- z = 2 (1 + it-^1 - 2 (1 + ir-^2 = 2 (1 + it-^2 [(1 + i) - 1]. This
last expression simplifies to 2i(l +it-^2 • Now, Zn= (1 +it. Since Logis a one-to-one function, the problem is solved by showing Log [(1 +it] =
Log [2i (1 + it-^2 ]. Use properties of the logarithm to do this.
- No. ia+ib = e"^2 "ncos(b2nn) + ie"^2 "nsin(b2nn), where n is an integer.
- The number c must be real, and WI = 1.
Sectio n 5.4. Trigonometric and H y p erbolic Functions: p a ge 186
d d [^00 n 2n ]^00 n (2n)z^2 n -l- dz COSZ = dz L; (- 1) (~n)! = L; (-1) (Zn)!. Explain why the in-
n=O n=l
dex n begins at 1 in the last expression. The result follows from simplification
and reindexing.