What is expand (z^n-e^{jθ})(z^n-e^{-jθ}) ?

The solution to expand (z^n-e^{jθ})(z^n-e^{-jθ}) is z^{2n}-e^{-jθ}z^n-e^{jθ}z^n+1

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expand (z^n-e^{jθ})(z^n-e^{-jθ})

expand

(z^{n} - e^{j θ}) (z^{n} - e^{- j θ})

z^{2 n} - e^{- j θ} z^{n} - e^{j θ} z^{n} + 1

Solution steps

(z^{n} - e^{j θ}) (z^{n} - e^{- j θ})

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = z^{n}, b = - e^{j θ}, c = z^{n}, d = - e^{- j θ}

= z^{n} z^{n} + z^{n} (- e^{- j θ}) + (- e^{j θ}) z^{n} + (- e^{j θ}) (- e^{- j θ})

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= z^{n} z^{n} - e^{- j θ} z^{n} - e^{j θ} z^{n} + e^{j θ} e^{- j θ}

Simplify

z^{n} z^{n} - e^{- j θ} z^{n} - e^{j θ} z^{n} + e^{j θ} e^{- j θ} : z^{2 n} - e^{- j θ} z^{n} - e^{j θ} z^{n} + 1

= z^{2 n} - e^{- j θ} z^{n} - e^{j θ} z^{n} + 1

\frac{3}{( 2 - z ) ^{2}}

(n - r)! (n + 1 - r)

\frac{3 x - 2}{x ( x - 2 ) ( x + 1 ) ( x + 3 )}

\frac{- x + 1}{5 ( x + 2 )}

2^{x (x - 1)}

What is expand (z^n-e^{jθ})(z^n-e^{-jθ}) ?
The solution to expand (z^n-e^{jθ})(z^n-e^{-jθ}) is z^{2n}-e^{-jθ}z^n-e^{jθ}z^n+1