What is e^{7+i(-pi/2+2kpi)} ?

The solution to e^{7+i(-pi/2+2kpi)} is e^7cos((-pi+4pik)/2)+e^7isin((-pi+4pik)/2)

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simplify e^{7+i(-pi/2+2kpi)}

simplify

e^{7 + i (- \frac{π}{2} + 2 kπ)}

e^{7} cos (\frac{- π + 4 πk}{2}) + e^{7} i sin (\frac{- π + 4 πk}{2})

Solution steps

e^{7 + i (- \frac{π}{2} + 2 kπ)}

Apply imaginary number rule:

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

= e^{7} (cos (\frac{4 πk - π}{2}) + i sin (\frac{4 πk - π}{2}))

Rewrite

e^{7} (cos (\frac{- π + 4 πk}{2}) + sin (\frac{- π + 4 πk}{2}) i)

in standard complex form:

e^{7} cos (\frac{- π + 4 πk}{2}) + e^{7} sin (\frac{- π + 4 πk}{2}) i

= e^{7} cos (\frac{- π + 4 πk}{2}) + e^{7} sin (\frac{- π + 4 πk}{2}) i

(m + p = 5, 5) (f (p) - p = 2, 7) (f (p) + m = 28)

\frac{2 z}{( 2 z - 1 ) ( 4 z - 1 )}

(4 + 6 i)^{8}

\frac{6 + s ^{5} - 4 s ^{4}}{s ^{4} ( s - 2 ) ^{2}}

\frac{x ^{2} + 4 x + 1}{( x - 1 ) ^{2} \cdot ( x ^{2} + 1 )}

What is e^{7+i(-pi/2+2kpi)} ?
The solution to e^{7+i(-pi/2+2kpi)} is e^7cos((-pi+4pik)/2)+e^7isin((-pi+4pik)/2)