Example 1 Show that
cosh z = cosh x cos y + i sinh x sin y
sinh z = sinh x cos y + i cosh x sin y
Write z = x + iy and have cosh z = cosh(x + iy). Then recall trigonometric identities
cosh(x + y) = cosh x cosh y + sinh x sinh y.
cosh(x + iy) = cosh x cosh(iy) + sinh x sinh(iy).
Since cosh(iy) = cos y and sinh(iy) = i sin y (we can easily show by writing them in
exponential forms),
cosh(x + iy) = cosh x cos(y) + i sinh x sin(y).
We do the same for sinh z.
sinh z = sinh(x + iy) = sinh x cosh(iy) + cosh x sinh(iy).
Since cosh(iy) = cos y and sinh(iy) = i sin y,
sinh(x + iy) = sinh x cos y + i cosh x sin y.
Example 2
Find all solutions of cos z = −1.
Write cos z =
eiz +e−iz
2
= −1. Then, by letting M = eiz , we have
M + M −1
= −1,
2
which yields the quadratic equation
M 2 + M + 2 = 0,
(M + 1)2 = 0.
The solution to the equation is M = −1. Since M = eiz ,
eiz = −1.
All we need to do is to take natural log to both sides, but before doing that, make sure to
write −1 in the exponential form with i2πn.
eiz = −1
= ei(π+2πn) ,
n = 0, ±1, ±2, . . .
By taking natural log,
iz = iπ + i2πn, n = 0, ±1, ±2, . . .
z = π + 2πn, n = 0, ±1, ±2, . . .
1
2
The answer was expectable because obviously cos(π + 2πn) = −1 for integers n. However, it is not so obvious often (for example, cos z = −i5), so make sure to understand the
way of solving as shown above.
Example 3 Find Lnz and ln z of z = 2 − i.
Write z = 2 − i in an exponential form. Since Arg(z) = −π/3,
√
z = 2 − i = 5ei(−π/3+2πn) , n = 0, ±1, ±2, . . .
Then,
√
π
ln z = ln 5 − i + i2πn, n = 0, ±1, ±2, . . .
3
The principal value Lnz corresponds to the value of ln z without i2πn so that
√
π
Lnz = ln 5 − i .
3