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690 CHAPTER 19 Complex Numbers and Functions

SECTION 19.4 PROBLEMS

In each of Problems 1 through 6, determine all values of 6. 5
the complex logarithm of z.
7. Let z and w be nonzero complex numbers. Show that
1. −4i each value of log(zw) is equal to a value of log(z) plus
a value of log(w).
2. 2 − 2i
8. Let z and w be nonzero complex numbers. Show that
3. −5
each value of log(z/w) is equal to a value of log(z)
4. 1 + 5i minus a value of log(w).
5. −9 + 2i

19.5 Powers

w
We want to assign a meaning to z when z and w are complex and z =0. If w is a positive integer,
n
3
n
z is clear. For example, z = z · z · z.And z −n = 1/z if z = 0. For other powers, we will proceed
in stages.
nth Roots
Let n be a positive integer. An nth root z 1/n of z is a number z 1/n whose nth power is z. We want
all values of z 1/n . To find these, begin with the polar form of z,

z =re i(θ+2kπ)
with all of the arguments θ + 2kπ of z in the exponent. Then
e
z 1/n =r 1/n i(θ+2kπ)/n
in which r 1/n is the real nth root of the positive number r.As k varies over the integers, the
numbers on the right give all the nth roots of z.
For k = 0,1,··· ,n − 1, we obtain n distinct numbers

e
e
e
r 1/n iθ/n , r 1/n i(θ+2π)/n , r 1/n i(θ+4π)/n ··· and r 1/n i(θ+2(n−1)π)/n . (19.8)
e
These are all nth roots of z. Other choices of k reproduce numbers already in this list. For
example, with k = n we get
r 1/n i(θ+2nπ)/n =r 1/n iθ/n 2πi =r 1/n iθ/n
e
e
e
e
because e 2πi = 1. Therefore k = n gives us the first number in the list (19.8) corresponding to
k = 0.
If k = n + 1, we obtain
e
r 1/n i(θ+2(n+1)π)/n =r 1/n i(θ+2π)/n 2πi =r 1/n i(θ+2π)/n ,
e
e
e
which is the second number in the list (19.8), corresponding to k = 1.
To sum up, the nth roots of z are the n numbers
r 1/n i(θ+2kπ)/n for k = 0,1,··· ,n − 1.
e
These can be written as

θ + 2kπ θ + 2kπ
1/n
r cos + i sin for k = 0,1,··· ,n − 1.
n n

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October 15, 2010 18:5 THM/NEIL Page-690 27410_19_ch19_p667-694