What is e^{t(-1/a+i)} ?

The solution to e^{t(-1/a+i)} is e^{-t/a}cos(t)+ie^{-t/a}sin(t)

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simplify e^{t(-1/a+i)}

simplify

e^{t (- \frac{1}{a} + i)}

e^{- \frac{t}{a}} cos (t) + i e^{- \frac{t}{a}} sin (t)

Solution steps

e^{t (- \frac{1}{a} + i)}

Apply imaginary number rule:

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

= e^{- \frac{t}{a}} (cos (t) + i sin (t))

Rewrite

e^{- \frac{t}{a}} (cos (t) + sin (t) i)

in standard complex form:

e^{- \frac{t}{a}} cos (t) + e^{- \frac{t}{a}} sin (t) i

= e^{- \frac{t}{a}} cos (t) + e^{- \frac{t}{a}} sin (t) i

\frac{- 6 x + 17}{( x + 3 ) ( 2 x - 1 )}

- \frac{x ^{4} - 10 x ^{3} - 13 x ^{2} + 60 x + 36}{( x ^{2} - 5 x + 6 ) ^{2}}

\frac{s + 1}{s ( s ^{2} - 7 s )}

2 x^{2} y sin (x^{2} + y^{2}) + 2 a x sin (a (x^{2} + y^{2}))

\frac{( ( t ^{4} ) - ( 10 t ^{\frac{3}{2}} ) )}{( t ^{\frac{9}{2}} )}

What is e^{t(-1/a+i)} ?
The solution to e^{t(-1/a+i)} is e^{-t/a}cos(t)+ie^{-t/a}sin(t)