What is e^{i(kz-ωt)} ?

The solution to e^{i(kz-ωt)} is cos(kz-ωt)+isin(kz-ωt)

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simplify e^{i(kz-ωt)}

simplify

e^{i (k z - ω t)}

cos (k z - ω t) + i sin (k z - ω t)

Solution steps

e^{i (k z - ω t)}

Apply imaginary number rule:

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

= cos (k z - ω t) + i sin (k z - ω t)

1 + ((3 ln (x)) - \frac{x ^{2}}{24})^{2}

\frac{( y ^{2} - x ^{2} ) y}{x ( y ^{2} + x ^{2} )}

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

\frac{d}{d x} ((x - 1) tan (x)) + x sin (x)

\frac{s ( 1 - e ^{- 2 π s} )}{s ^{2} + 4}