1复杂变量-简介/修订。

z =reiθ= r（cosθ+isinθ），（1.1）

ez = ez̄，sinz = sinz，z3 = z3，（1 + i）z =（1 − i）z̄=（1 + i）z̄。

（1.2）

（1.3）

（1.4）

（1.5）（1.6）

（1.7）

f′（z0）= limΔz→0

lim（z0 + ∆z）（z̄0+ ∆z）− z0z̄0

f（z0 +Δz）− f（z0），Δz

f（z0）= lim ∆z→0

（z 0 + ∆ z）2 − z 02 ∆z

Δz→0

Δz

Variables

1 Complex variables – introduction/revision.
Complex numbers. We write z = x+iy, where x = Re(z),y = Im(z) are the real and imaginary parts of z.

Alternatively, using polar coordinates in the complex plane z,
z
=re=r(cosθ+isinθ), (1.1)

where r = |z| = 􏰅x2 + y2 is the absolute value (magnitude, modulus) of z and θ = arg(z) is the argument of z. The argument is an example of a multi-valued function since exp[i(θ + 2πn)] = exp() for any integer n. When a single- valued quantity is required we can choose the principal value of the argument, Arg(z) such that Arg(z) (π, π], for example. In e􏰀ect, this creates a branch cut in the complex plane z along the negative real axis with the points z = x < 0 included on the upper side of the cut.

Exercise. Find |z| and Arg(z) for the following complex numbers: z = 1+i, z = 1i, z = 1i, z = 1+i. Note that the ‘standard’ formula for the polar angle θ = arctan(y/x) does not always work in these calculations.

The complex conjugate of a complex number z = x + iy is z ̄ = x iy. Exercise. Verify that |z|2 = zz ̄.

Functions. We write, for example, w = f(z) = u(x,y)+iv(x,y). For many standard functions we have f(z) = f(z ̄), for example

however

ez = ez ̄,sinz = sinz,z3 = z3, (1 + i)z = (1 i)z ̄ ̸= (1 + i)z ̄.

(1.2)

(1.3)

(1.4)

(1.5) (1.6)

(1.7)

Clearly the rule f(z) = f(z) applies if the function can be written as a Taylor series with real coe􏰂cients. Di􏰀erentiation. The function f(z) is di􏰀erentiable at some point z0 if the limit,

f(z0) = lim z0

exists and does not depend on how z tends to zero. Example. Let f(z) = z2. We have

which is a familiar derivative of z2.
Example. Let f(z) = |z|2. Write f(z) = zz ̄ and observe that

lim (z0 + ∆z)(z ̄0 + ∆z) z0z ̄0

f(z0 + ∆z) f(z0), z

f(z0)= lim z0

( z 0 + ∆ z ) 2 z 02 z

z0

z

= lim(2z0+∆z)=2z0, z0

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