how do I solve cosh((1+j)/2) to form of a+jb????
cosh(x+jy) = cosh x cos y + j sinh x sin y
here x = 1/2
and y = 1/2
To solve cosh((1+j)/2) and express it in the form a + jb, we need to understand that cosh(x) is defined as the sum of the exponential function e^x and its inverse e^(-x) divided by 2.
Let's break down the steps to solve this:
1. Substitute the given value into the cosh function:
cosh((1+j)/2)
2. Express this value using exponential functions:
cosh((1+j)/2) = (e^((1+j)/2) + e^(-(1+j)/2))/2
3. Simplify the exponents individually:
e^((1+j)/2) = e^(1/2) * e^(j/2)
e^(-(1+j)/2) = e^(-1/2) * e^(-j/2)
Remember that e^(j/2) can be expressed as cos(j/2) + j * sin(j/2), by Euler's formula: e^(jθ) = cos(θ) + j * sin(θ).
4. Substitute the values obtained from the previous step back into the cosh equation:
cosh((1+j)/2) = (e^(1/2) * (cos(j/2) + j * sin(j/2)) + e^(-1/2) * (cos(-j/2) + j * sin(-j/2)))/2
5. Simplify further by combining like terms:
cosh((1+j)/2) = (e^(1/2) * cos(j/2) + e^(-1/2) * cos(-j/2))/2 + (e^(1/2) * j * sin(j/2) + e^(-1/2) * j * sin(-j/2))/2
6. Now, let's work on the trigonometric functions with complex values:
cos(-j/2) = cos(j/2) (since cosine is an even function)
sin(-j/2) = -sin(j/2) (since sine is an odd function)
7. Substitute the trigonometric values back into the expression:
cosh((1+j)/2) = (e^(1/2) * cos(j/2) + e^(-1/2) * cos(j/2))/2 + (e^(1/2) * j * sin(j/2) - e^(-1/2) * j * sin(j/2))/2
8. Combine like terms again:
cosh((1+j)/2) = [(e^(1/2) + e^(-1/2)) * cos(j/2)]/2 + [(e^(1/2) - e^(-1/2)) * j * sin(j/2)]/2
Let's define:
a = (e^(1/2) + e^(-1/2))/2
b = (e^(1/2) - e^(-1/2))/2
9. Substitute the values of a and b back into the expression:
cosh((1+j)/2) = a * cos(j/2) + b * j * sin(j/2)
Therefore, cosh((1+j)/2) in the form a + jb is:
cosh((1+j)/2) = a * cos(j/2) + b * j * sin(j/2)