What is (e^{-2it}+1)*te^{it} ?

The solution to (e^{-2it}+1)*te^{it} is t(cos(2t)-isin(2t)+1)(cos(t)+isin(t))

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simplify (e^{-2it}+1)*te^{it}

simplify

(e^{- 2 i t} + 1) \cdot t e^{i t}

t (cos (2 t) - i sin (2 t) + 1) (cos (t) + i sin (t))

Solution steps

(e^{- 2 i t} + 1) t e^{i t}

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

= e^{i t} t (cos (2 t) - i sin (2 t) + 1)

e^{i t} = cos (t) + i sin (t)

= t (cos (2 t) - i sin (2 t) + 1) (cos (t) + i sin (t))

\frac{4}{( y + 1 ) ( y + 5 )}

\frac{3 ( 3 x + 4 )}{( x - 4 ) ( x + 4 )}

(\frac{- 36}{9} bi g)^{2} (9 - 36) 2

\frac{( x + y )}{( x - y ) ( x - 2 y )}

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

What is (e^{-2it}+1)*te^{it} ?
The solution to (e^{-2it}+1)*te^{it} is t(cos(2t)-isin(2t)+1)(cos(t)+isin(t))