Math 455      Soln to Selected Problem from Sec. 35.
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1. (#11, page 136)

   sinz = cos z => tan z=1 => z = tan^(-1) 1

   Hence by equation (10) in the text

   z = (i/2) log ((1-i)/(1+i)) = (i/2) log(-i) = (i/2)(-i\pi/2 + 2k\pi i)
     = \pi/4 + k \pi
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2. (#15b, p.136)

   First try the principal branch of the log.

   (4 + z^2)^(1/2) = e^[(1/2) Log(4+z^2)]

   Let z = x + iy. Then,

   4+z^2 = (4+x^2-y^2) + i(2xy).

   x=0 => for y\geq 2 or y \leq -2 we would have 4-y^2 \leq 0, and

   y=0 => 4+x^2 \leq 0 does not havea soln.

   Hence, the principal branch of log would give us a branch of (4+z^2)^(1/2)
   that is analytic in any region of the complex plain that contains the
   slit along the imaginary axis from -2i to 2i. This is what we wanted.

   Note that if you factor z^2 form the given function and then use the
   principal branch of the log, the resulting branch of (4+z^2)^(1/2)
   would be analytic in the complement of the slit above.
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