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 =

n=0
(−1)n
x2n+1
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 =

n=0
(−1)n
z2n+1
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
eπi/4, e3πi/2,
and
e13πi/6
in standard form.
3. Express 1/

2 + i/

2 in the form
ez
for some complex number z.
4. Find all values of z for which
ez
= 1 +

3 i.
5. Prove that
ez
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
1.3.7).
7. Verify that ez+2πi = ez for every z C (Part (d) of Theorem 1.3.7).
8. Verify that
ez
= 1 if and only if z = 2πni for some interger n (Part (e) of
Theorem 1.3.7).
9. For which values of z is
ez
a real number? For which values is it an imaginary
number?
10. Show that if the real power series
∑∞
n=0
anxn∑∞
has radius of convergence R,
then the corresponding complex power series
n=0
anzn
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
1.3.4).
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.
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