Tuesday, January 1, 2013

Complex Analysis: #1 Complex Numbers

  • Complex Analysis: #1 Complex Numbers

An “imaginary” number is introduced, called i (for imaginary)(sometimes j is also used in physics), which is declared to be a solution of the polynomial equation

x2+ 1 = 0.

The field of complex numbers is denoted by ℂ. We have

ℂ = {x + iy : x, y ∈ ℜ}. 

For z = x + iy, we also write Re(z) to denote the real part of z, namely the real number x. Also Im(z) = y is the imaginary part of z.

Let z1 = x1 + iy1 and z2 = x2 + iy2 be complex numbers. Then the addition and multiplication operations are given by

 z1 + z2 = (x1 + x2) + i(y1 + y2)
and
z1 · z2 = (x1·x2 − y1·y2) + i(x1·y2 − x2·y1).

The complex conjugate is z = x − iy (that is, z = x+i(−y)). Therefore the complex conjugate of z = z Complex Analysis: #1 Complex Numbers complex conjugate conjugate. We have |z|2 = zz = x2 + y2. If z ≠ 0 (that is, either x ≠ 0, or y ≠ 0) then zz > 0, and we have

Complex Analysis: #1 Complex Numbers equation pic 1
If z ≠ 0 then there are unique real numbers r > 0 and 0 ≤ θ < 2π such that x = r cos θ and y = r sin θ. So let u, v ∈ ℂ be non-zero numbers, and let

u = r(cos θ + i sin θ), 
v = s(cos ψ + i sin ψ). 

Then (remembering the rules for combining trigonometric functions), we see that

u · v = r · s(cos(θ + ψ) + i sin(θ + ψ)). 

If ℂ is identified with ℜ2, the 2-dimensional real vector space, then we can identify any complex number z = x +iy. But for any two real numbers x and y, there exists a unique pair of numbers r ≥ 0, 0 ≤ θ < 2π, with x = r cos θ and y = r sin θ. (Obviously it is not quite unique for the number z = 0) So let u = s + it be some other complex number. Then

Complex Analysis: #1 Complex Numbers equation pic 2

Thus we see that multiplication of complex numbers looks like an (orientation preserving) orthogonal mapping within ℜ2 — combined with a scalar factor r.

More generally, let f : ℜ2 → ℜ2 be an arbitrary linear mapping, represented by the matrix
Complex Analysis: #1 Complex Numbers equation pic 3
 How can this mapping be represented in terms of complex arithmetic? We have
Complex Analysis: #1 Complex Numbers equation pic 4
Therefore, using the linearity of f, for z = x + iy we have

Complex Analysis: #1 Complex Numbers equation pic 5

Therefore, if f(1) = −if(i), that is, if(1) = f(i), then the mapping is simply complex multiplication. On the other hand, if f(1) = if(i) then we have f(z) = w · z, where w = (f(1) + if(i))/2 ∈ ℂ is some complex number.

Complex Analysis: #1 Complex Numbers equation pic 6

for a suitable choice of r and ψ. This is an orientation reversing rotation (again combined with a scalar factor r).

Now let u = a + ib and v = c + id. What is the scalar product <u, v>? It is

Complex Analysis: #1 Complex Numbers equation pic 7

Therefore <z, z> = |z|2, where |z| = √x2 + y2 is the absolute value of z = x + iy.

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