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.