What is e^{-jt}(e^{t(j-k)}-1) ?

The solution to e^{-jt}(e^{t(j-k)}-1) is (e^{t(-k+j)}-1)/(e^{jt)}

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simplify e^{-jt}(e^{t(j-k)}-1)

simplify

e^{- j t} (e^{t (j - k)} - 1)

\frac{e ^{t (- k + j)} - 1}{e ^{j t}}

Solution steps

e^{- j t} (e^{t (j - k)} - 1)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{j t}} (e^{t (- k + j)} - 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( e ^{t (j - k)} - 1 )}{e ^{j t}}

1 \cdot (e^{t (j - k)} - 1) = e^{t (j - k)} - 1

= \frac{e ^{t (- k + j)} - 1}{e ^{j t}}

\frac{x - 2}{3 - x}

e^{3 x} \cdot 3 e^{x}

a t^{b} e^{- \frac{t}{c}}

e^{2 x} \cdot 2 e^{\frac{1}{2} x}

- 2 e^{- 2 y} ln (3 + x)

What is e^{-jt}(e^{t(j-k)}-1) ?
The solution to e^{-jt}(e^{t(j-k)}-1) is (e^{t(-k+j)}-1)/(e^{jt)}