What is (t-e^{iθ})(t-2) ?

The solution to (t-e^{iθ})(t-2) is (t^2-2t-tcos(θ)+2cos(θ))+i(2sin(θ)-tsin(θ))

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simplify (t-e^{iθ})(t-2)

simplify

(t - e^{i θ}) (t - 2)

(t^{2} - 2 t - t cos (θ) + 2 cos (θ)) + i (2 sin (θ) - t sin (θ))

Solution steps

(t - e^{i θ}) (t - 2)

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (t - (cos (θ) + i sin (θ))) (t - 2)

Expand

t - (cos (θ) + sin (θ) i) : t - cos (θ) - sin (θ) i

= (- cos (θ) - i sin (θ) + t) (t - 2)

Rewrite

(t - cos (θ) - sin (θ) i) (t - 2)

in standard complex form:

(t^{2} - 2 t - t cos (θ) + 2 cos (θ)) + (- t sin (θ) + 2 sin (θ)) i

= (t^{2} - 2 t - t cos (θ) + 2 cos (θ)) + (- t sin (θ) + 2 sin (θ)) i

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

\frac{24 x ( x - 2 )}{( 8 x ^{2} - 2 x - 1 ) z ( z - 2 )}

\frac{5 ( 2 x + 1 )}{( x - 7 ) ( x + 3 )}

5 sin (4 (x + 3)) + 7

(1 + x) \times cos (w x)

What is (t-e^{iθ})(t-2) ?
The solution to (t-e^{iθ})(t-2) is (t^2-2t-tcos(θ)+2cos(θ))+i(2sin(θ)-tsin(θ))