What is e^{2(1-3i)z}+e^{2(1+3i)z} ?

The solution to e^{2(1-3i)z}+e^{2(1+3i)z} is 2e^{2z}cos(6z)

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simplify e^{2(1-3i)z}+e^{2(1+3i)z}

simplify

e^{2 (1 - 3 i) z} + e^{2 (1 + 3 i) z}

2 e^{2 z} cos (6 z)

Solution steps

e^{2 (1 - 3 i) z} + e^{2 (1 + 3 i) z}

e^{2 (1 - 3 i) z} = e^{2 z} (cos (6 z) - i sin (6 z))

e^{2 (1 + 3 i) z} = e^{2 z} (cos (6 z) + i sin (6 z))

= e^{2 z} (cos (6 z) - i sin (6 z)) + e^{2 z} (cos (6 z) + i sin (6 z))

= e^{2 z} (cos (6 z) - i sin (6 z)) + e^{2 z} (cos (6 z) + i sin (6 z))

Expand

e^{2 z} (cos (6 z) - i sin (6 z)) : e^{2 z} cos (6 z) - i e^{2 z} sin (6 z)

= e^{2 z} cos (6 z) - i e^{2 z} sin (6 z) + e^{2 z} (cos (6 z) + sin (6 z) i)

Expand

e^{2 z} (cos (6 z) + sin (6 z) i) : e^{2 z} cos (6 z) + i e^{2 z} sin (6 z)

= e^{2 z} cos (6 z) - i e^{2 z} sin (6 z) + e^{2 z} cos (6 z) + i e^{2 z} sin (6 z)

Simplify

e^{2 z} cos (6 z) - i e^{2 z} sin (6 z) + e^{2 z} cos (6 z) + i e^{2 z} sin (6 z) : 2 e^{2 z} cos (6 z)

= 2 e^{2 z} cos (6 z)

(- 1, 2) y (2, - 3)

(1998 - 400 \cdot ln (\frac{400}{25})) - (1998 - 400 \cdot ln (20))

\frac{( 2 x - 3 ) ^{3}}{8}

\frac{a ( 2 x + 1 ) + 2}{a ( - x w - 1 + w )}

\frac{( 2 x - 3 ) ^{3}}{6}

What is e^{2(1-3i)z}+e^{2(1+3i)z} ?
The solution to e^{2(1-3i)z}+e^{2(1+3i)z} is 2e^{2z}cos(6z)