What is e^{2(a+bi)} ?

The solution to e^{2(a+bi)} is e^{2a}cos(2b)+ie^{2a}sin(2b)

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simplify e^{2(a+bi)}

simplify

e^{2 (a + bi)}

e^{2 a} cos (2 b) + i e^{2 a} sin (2 b)

Solution steps

e^{2 (a + bi)}

Apply imaginary number rule:

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

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

Rewrite

e^{2 a} (cos (2 b) + sin (2 b) i)

in standard complex form:

e^{2 a} cos (2 b) + e^{2 a} sin (2 b) i

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

\frac{5 - 2 ( a + 3 b )}{2}

(3 e^{5 t} - 3 e^{4 t}) (15 e^{5 t} - 12 e^{4 t})

3 sec (\frac{π}{4} (x + 1))

ln (cos^{2} (t) + 1) \cdot ((e^{x^{2}} \cdot 2 x) i + sinh (t) j)

e^{2 x} (x^{3} + 5)

What is e^{2(a+bi)} ?
The solution to e^{2(a+bi)} is e^{2a}cos(2b)+ie^{2a}sin(2b)