Complex Analysis - Maths KU

Note: ln z will be used to denote the
complex logarithm.
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4.1 Exponential and Logarithmic Functions 183
Hereafter, the notation ln z will alwaysbe used to denote the multiplevalued
complex logarithm. By switching to exponential notation z = re iθ in
(11), we obtain the following alternative description of the complex logarithm:
ln z = log e r + i(θ +2nπ), n=0, ±1, ±2,.... (12)
From (10) we see that the complex logarithm can be used to find all
solutions to the exponential equation e w = z when z isa nonzero complex
number.
EXAMPLE 3 Solving Exponential Equations
Find all complex solutions to each of the following equations.
(a) e w = i (b) e w =1+i (c) e w = −2
Solution For each equation e w = z, the set of solutions is given by w =lnz
where ln z isfound using Definition 4.2.
(a) For z = i, wehave|z| = 1 and arg(z) =π/2+2nπ. Thus, from (11) we
obtain:
�
π
w =lni = loge 1+i
2 +2nπ
�
.
Since log e 1 = 0, thissimplifiesto:
(4n +1)π
w = i, n =0, ±1, ±2 ... .
2
Therefore, each of the values:
w = ..., − 7π
2
i, −3π
2
i, π
2
5π
i, i, ...
2
satisfies the equation e w = i. The solutions to the equation e w = i given
on page 182 correspond to the values of ln i for n =0,1,and−1.
(b) For z =1+i, we have |z| = √ 2 and arg(z) =π/4+2nπ. Thus, from (11)
we obtain:
√ �
π
w =ln(1+i) = loge 2+i
4 +2nπ
�
.
√
1
Because loge 2= 2 loge 2, thiscan be rewritten as:
w = 1
2 log e 2+
Each value of w isa solution to e w =1+i.
(8n +1)π
i, n =0, ±1, ±2,... .
4
(c) Again we use (11). Since z = −2, we have |z| = 2 and arg(z) =π +2nπ,
and so:
w = ln(−2) = log e 2+i (π +2nπ) .