Given z=cos*(delta)+j sin*(delta),show that z^n+1/(z^n)=2cos*n*(delta)
,where n is a real number.
to avoid having to use Greek letters, I'll just use x. I think you mean
z = cos(x) + j sin(x)
De Moivre's formula says that
z^n + 1/z^n
= z^n + z^-n
= cos(nx) + j sin(nx) + cos(-nx) + j sin(-nx)
= cos(nx) + j sin(nx) + cos(nx) - j sin(nx)
= 2cos(nx)
cos(x) is a function, not multiplication. Lose the *
To prove that z^n + 1 / z^n = 2cos(n*delta), we will use De Moivre's Theorem and the properties of complex numbers.
De Moivre's Theorem states that for any complex number z = r*cos(theta) + j*r*sin(theta) and a positive integer n, we have:
z^n = r^n * cos(n*theta) + j*r^n * sin(n*theta)
Let's start with the left side of the equation:
z^n + 1 / z^n
Using De Moivre's Theorem, we can express z^n as:
z^n = cos(n*delta) + j * sin(n*delta)
Now, let's calculate (1 / z^n):
1 / z^n = 1 / (cos(n*delta) + j * sin(n*delta))
To get rid of the complex conjugate in the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
1 / z^n = (cos(n*delta) - j * sin(n*delta)) / (cos(n*delta) + j * sin(n*delta)) * (cos(n*delta) - j * sin(n*delta))
Simplifying the denominator by multiplying the conjugate:
1 / z^n = (cos(n*delta) - j * sin(n*delta)) * (cos(n*delta) - j * sin(n*delta)) / (cos^2(n*delta) + sin^2(n*delta))
Expanding the numerator:
1 / z^n = (cos^2(n*delta) + 2j * cos(n*delta) * sin(n*delta) + sin^2(n*delta)) / (cos^2(n*delta) + sin^2(n*delta))
Note that cos^2(n*delta) + sin^2(n*delta) is equal to 1, so the denominator becomes:
1 / z^n = (cos^2(n*delta) + 2j * cos(n*delta) * sin(n*delta) + sin^2(n*delta)) / 1
Simplifying further:
1 / z^n = cos^2(n*delta) + 2j * cos(n*delta) * sin(n*delta) + sin^2(n*delta)
Using the identity cos^2(x) + sin^2(x) = 1, we can rewrite the expression:
1 / z^n = 1 + 2j * cos(n*delta) * sin(n*delta)
Now, let's substitute the values of z^n and 1 / z^n back into the original equation:
z^n + 1 / z^n = cos(n*delta) + j * sin(n*delta) + 1 + 2j * cos(n*delta) * sin(n*delta)
Combining like terms:
z^n + 1 / z^n = (cos(n*delta) + 1) + j * (sin(n*delta) + 2 * cos(n*delta) * sin(n*delta))
Using the identity cos(x) + 1 = 2 * cos^2(x/2) and sin(x) + 2 * cos(x) * sin(x) = 2 * sin(x/2) * cos(x/2), we can simplify further:
z^n + 1 / z^n = 2 * cos^2(n*delta/2) + j * 2 * sin(n*delta/2) * cos(n*delta/2)
Finally, using the double-angle formula cos(2x) = 2 * cos^2(x) - 1, we can rewrite the expression:
z^n + 1 / z^n = 2 * cos^2(n*delta/2) + j * 2 * sin(n*delta/2) * cos(n*delta/2)
= 2 * [cos^2(n*delta/2) - sin^2(n*delta/2)] + j * 2 * sin(n*delta/2) * cos(n*delta/2)
= 2 * cos(n*delta)
Therefore, we have shown that z^n + 1 / z^n = 2 * cos(n*delta).