What is e^{i(θ+2kpi)} ?

The solution to e^{i(θ+2kpi)} is cos(θ+2pik)+isin(θ+2pik)

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simplify e^{i(θ+2kpi)}

simplify

e^{i (θ + 2 kπ)}

cos (θ + 2 πk) + i sin (θ + 2 πk)

Solution steps

e^{i (θ + 2 kπ)}

Apply imaginary number rule:

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

= cos (θ + 2 πk) + i sin (θ + 2 πk)

\frac{( 1 - 3 x ) ( 1 + 3 x )}{9 x ^{2}}

d (\frac{f ( x )}{w ( t - 1 )})

\frac{1}{( s - j + 1 ) ( s + j + 1 )}

11 (π + 2)

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

What is e^{i(θ+2kpi)} ?
The solution to e^{i(θ+2kpi)} is cos(θ+2pik)+isin(θ+2pik)