In and , complex random variables are a generalization of real-valued to , i.e. the possible values a complex random variable may take are complex numbers. Complex random variables can always be considered as pairs of real random variables: their real and imaginary parts. Therefore, the of one complex random variable may be interpreted as the of two real random variables.
Some concepts of real random variables have a straightforward generalization to complex random variables—e.g., the definition of the of a complex random variable. Other concepts are unique to complex random variables.
* A complex random variable Z on the (is a on Z: ,such that both its real part R(z) and its imaginary part i(z) are real on ..
Circular symmetric complex normal random variables are used extensively in signal processing, and are sometimes referred to as just complex normal in signal processing literature. Formally, is a standard complex normal random variable.
The characteristic function of complex normal distribution is given by is a n -dimensional complex vector. Central limit theorem. If. The modulus of a complex normal random variable follows a Hoyt distribution
Example:1 Find fourth root of
Solution: Convert to polar first: r=
Example:2 Find the fifth root of 4-4i of
Solution: Convert 4-4i to polar: r==;()cis()
Example1: Solve cosh2 x – sinh2 x
Given: cosh2 x – sinh2 x
We know that
Sinh x = [ex– e-x]/2
Cosh x = [ex + e-x]/2
Cosh2 x – sinh2 x = [ [ex + e-x]/2 ]2 – [ [ex – e-x]/2 ]2
Cosh2 x – sinh2 x = (4ex-x) /4
Cosh2 x – sinh2 x = (4e0) /4
Cosh2 x – sinh2 x = 4(1) /4 = 1
Therefore, cosh2 x – sinh2 x = 1
A ray through the unit hyperbola in the point , where is twice the area between the ray, the hyperbola, and the -axis
Find the inverse of the function f(x) = ln(x – 2)
First, replace f(x) with y
So, y = ln(x – 2)
Replace the equation in exponential way , x – 2 = ey
Now, solving for x,
x = 2 + ey
Now, replace x with y and thus, f-1(x) = y = 2 + ey
To solve an equation: f(x) = 2x + 3, at x = 4
f(4) = 2 × 4 + 3
f(4) = 11
Now, let’s apply for reverse on 11.
f-1(11) = (11 – 3) / 2
f-1(11) = 4
Magically we get 4 again.
Therefore, f(f(4)) = f(4)
So, when we apply function f and its reverse f-1 gives the original value back again, i.e, f-1(f(x)) = x.
Example1: Prove that function is analytic function.
Solution. Real and Imaginary parts of are
On differentiating u,v we get
Example2: Determine P such that the function analytic
f(z) is analytic Cauchy Riemann should be satisfied that is
Problem: simplify 16 i+10i (3-i)
16i +10i (3-i)
=16i+10i (3i) +10 i(-i)
Here real part is 10 and imaginary part is 46
Problem: express the following into a+ib form
z = = = + i
Modulus , = = =
Conjugate = ( -
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