What is e^{(2+i)t}-e^{(2-i)t} ?

The solution to e^{(2+i)t}-e^{(2-i)t} is 2ie^{2t}sin(t)

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

simplify

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

2 i e^{2 t} sin (t)

Solution steps

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

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

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

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

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

Expand

e^{2 t} (cos (t) + sin (t) i) : e^{2 t} cos (t) + i e^{2 t} sin (t)

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

Expand

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

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

Simplify

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

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

\frac{2 ( 5 x ^{2} - 3 x + 2 )}{x + 4}

1 + \frac{e ^{s} ( 1 - e ^{- s} )}{s}

R e (x + i y)^{4}

1 - 6 \cdot (y + 3 = - 23)

\frac{( a + b ) a}{x ( x + z )}

What is e^{(2+i)t}-e^{(2-i)t} ?
The solution to e^{(2+i)t}-e^{(2-i)t} is 2ie^{2t}sin(t)