18 1. The Complex Numbers
Example 1.3.9. How would you define a complex version of the arctan function?
Solution: The power series which converges to arctan x on (−1, 1) is
arctan x =
2n + 1
If we replace x by the complex variable z, we obtain a series which converges on
the disc D1(0). We then use it to define arctan z on this disc:
arctan z =
2n + 1
Exercise Set 1.3
1. Using the power series for ez, prove that ez = ez for each z ∈ C.
2. Express the complex numbers
in standard form.
3. Express 1/
2 + i/
2 in the form
for some complex number z.
4. Find all values of z for which
= 1 +
5. Prove that
is never 0 (Part (a) of Theorem 1.3.7).
6. Prove that | ez | = eRe(z) ≤ e|z| for each z ∈ C (Parts (b) and (c) of Theorem
7. Verify that ez+2πi = ez for every z ∈ C (Part (d) of Theorem 1.3.7).
8. Verify that
= 1 if and only if z = 2πni for some interger n (Part (e) of
9. For which values of z is
a real number? For which values is it an imaginary
10. Show that if the real power series
has radius of convergence R,
then the corresponding complex power series
also has radius of
convergence R (you may assume radius of convergence for a complex power
series makes sense and has the properties described in Remark 1.2.10).
11. Use induction on n to prove the complex binomial formula (1.3.1).
12. Derive the trigonometric identities
sin(z + w) = sin z cos w + cos z sin w,
cos(z + w) = cos z cos w − sin z sin w
from the law of exponents (Theorem 1.3.2) and Euler’s identity (Theorem
13. Show that cos ix = cosh x and sin ix = i sinh x, where cosh and sinh are the
hyperbolic cosine and hyperbolic sine from calculus.
14. Show that
sin(x + iy) = sin x cosh y + i cos x sinh y and
cos(x + iy) = cos x cosh y − i sin x sinh y.