What is e^{pi(1+i)} ?

The solution to e^{pi(1+i)} is -e^pi

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simplify e^{pi(1+i)}

simplify

e^{π (1 + i)}

- e^{π}

Decimal Notation

- 23.14069 \dots

Solution steps

e^{π (1 + i)}

Apply imaginary number rule:

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

= e^{π} (cos (π) + i sin (π))

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= e^{π} (- 1 + i sin (π))

sin (π) i = 0

= e^{π} (0 - 1)

Simplify

= - e^{π}

y^{2} (\frac{y ^{\frac{3}{2}}}{3} - \frac{y ^{3}}{24})

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- \frac{75 ( - 1 ) ^{\frac{1}{4}} ( 74 + 69 i )}{7754 - 5497 i}

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(2 + x) 9 - x^{2}