Advanced Engineering Math II phần 2 docx

11

2. We calculate w = ln z as follows: First write z in the form z = re

iø. Now let w = u+iv.

Then

e

w

= z

gives e

u+iv

= z = re

iø.

Thus, e

u

e

iv

= re

iø.

Equating magnitudes and arguments,

e

u

= r, v = ø,

or u = ln r, v = ø.

Thus, Formula for ln z

ln!z"="ln!r"+"iø,""r"="|z|,"ø"="arg(z)

3. If ø is chosen as the principal value of arg(z), that is, -π < ø ≤ π, then we get the

principal value of ln z, called Ln z. Thus,

Formula for Ln z

Ln!z"="ln!r"+"iø,"r"="|z|,"ø"="Arg(z)

Also ln!z"="Ln!z"+"i(2nπ);"n"="0,"±1,"±2,"...

What about the domain of the function Ln?

Answer: Ln: CI!!-{0}ÆCI!!. However, Ln is discontinuous everywhere along the negative

x-axis (where Arg(z) switches from π to numbers close to -π. If we want to make the Ln

continuous, we remove that nasty piece from the domain and take

Ln: {z | z ≠ 0 and arg(z) ≠ π} Æ CI!!

4. ln 0 is still undefined, as there is no complex number w such that e

w = 0.

Examples 3.7

(a) ln1 = 0 + 2nπi = 2nπi; Ln 1 = 0;

(b) ln4 = 1.386... + 2nπi; Ln 4 = 1.386...

(c) If r is real, then

ln r = the usual value of ln r + 2nπi; Ln r = ln r

(d) lni = πi/2 + 2nπi; Ln i = πi/2;

(e) ln(-1) = πi + 2nπi; Ln (-1) = πi;

(f) ln(3-4i) = ln5 + i arg(3-4i) + 2nπi

= ln5 + i arctan(4/3) + 2nπi; Ln(3-4i) = ln5 + i arctan(4/3).

More Properties

1. ln(z w) = lnz + lnw; ln(z/w) = ln(z) - ln(w).

This doesn't work for Ln; eg., z = w = -1 gives

Ln z + Ln w = πi + πi = 2πi,

but Ln(zw) = Ln(1) = 0.

2. Ln z jumps every time you cross the negative x-axis, but is continuous everywhere

else (except zero of course). If you want it to remain continuous, you must switch to

another branch of the logarithm. (Lnz is called the principal branch of the logarithm.)

3. e

ln z = z, and ln(e

z

) = z + 2nπi;

e

Ln z = z, and Ln(e

z

) = z + 2nπi;

(For example, z = 3πi gives e

z

= -1, and Ln(ez

) = πi ≠ z.)