Solutions and Remarks for Homework                Section 1.4, page 31 
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#5. |e^z| = |e^(x+iy)| = e^x |e^(iy)| = e^x, since e^x is a real number
    |e^x|=e^x and |e^(iy)|=|cos(y)+i sin(y)|=1.

    arg(e^z) = tan^(-1)(sin(y)/cos(y)) + 2k pi = tan^(-1)tan(y)=2kpi
    =y+2kpi.

#17.  Take an arbitrary point on the unit circle.  Then, the coordinates
      of this point are  z=(cos t, sin t)=cos t + i sin t = e^(it)
      for some central angle t \in [0, 2pi].

#20.  In this problem the following identities are needed:

      a. sin^(2) u = (1 - cos 2u) / 2

      b. sin u = (e^(iu) - e^(-iu)) / 2i which implies that

      c. sin u = e^(iu) (1 - e^(2iu)) / 2i

      d. sin u = e^(-iu) (e^(2iu) - 1) / 2i

      e. cos u = [e^(iu) + e^(-iu)/2

      We need to compute the real part of [1-e^i(n+1)t]/[1-e^(it)].
      To do so, we have to multiply the denominator by its
      conjugate [1 - e^(-it)]. After doing so and simplifying the
      denominator of the product will be 2(1-cos t) which, using
      identity (a) above with 2u=t) is equivalent to 4 sin^2 (t/2).
      Hence at this point you ahould have

      [1 - e^i(n+1)t] [1 - e^(-it)] / 4 sin^2 (t/2).

      Foil and simplify the numerator to get

      [(1 - e^(-it)) + e^(int) (1 - e^(it))] / 4 sin^2 (t/2)

      Now if you use identities (c) and (d) above with u=t, and
      simplify you will get the desired result.