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

The solution to i(e^{2+3i}-e^{-2-3i}) is (-sin(3)-e^4sin(3))/(e^2)+i(-cos(3)+e^4cos(3))/(e^2)

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

simplify

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

\frac{- sin ( 3 ) - e ^{4} sin ( 3 )}{e ^{2}} + i \frac{- cos ( 3 ) + e ^{4} cos ( 3 )}{e ^{2}}

Solution steps

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

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

e^{- 2 - 3 i} = \frac{1}{e ^{2}} (cos (3) - i sin (3))

= i (e^{2} (cos (3) + i sin (3)) - \frac{1}{e ^{2}} (- i sin (3) + cos (3)))

Apply complex arithmetic rule:

(ai) (b + c i) = - a c + abi

a = 1, b = \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}}, c = \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}}

= - 1 \cdot \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}} + 1 \cdot \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} i

- 1 \cdot \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}} = - \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}}

1 \cdot \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} = \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}}

= - \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}} + \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} i

Multiply

\frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} i : \frac{i ( e ^{4} cos ( 3 ) - cos ( 3 ) )}{e ^{2}}

= - \frac{e ^{4} sin ( 3 ) + sin ( 3 )}{e ^{2}} + \frac{i ( e ^{4} cos ( 3 ) - cos ( 3 ) )}{e ^{2}}

Apply rule

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

= \frac{- ( e ^{4} sin ( 3 ) + sin ( 3 ) ) + i ( e ^{4} cos ( 3 ) - cos ( 3 ) )}{e ^{2}}

Expand

- (e^{4} sin (3) + sin (3)) + (e^{4} cos (3) - cos (3)) i : - e^{4} sin (3) - sin (3) + e^{4} i cos (3) - i cos (3)

= \frac{- e ^{4} sin ( 3 ) - sin ( 3 ) + e ^{4} i cos ( 3 ) - i cos ( 3 )}{e ^{2}}

Rewrite

\frac{- e ^{4} sin ( 3 ) - sin ( 3 ) + e ^{4} i cos ( 3 ) - i cos ( 3 )}{e ^{2}}

in standard complex form:

\frac{- e ^{4} sin ( 3 ) - sin ( 3 )}{e ^{2}} + \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} i

= \frac{- e ^{4} sin ( 3 ) - sin ( 3 )}{e ^{2}} + \frac{e ^{4} cos ( 3 ) - cos ( 3 )}{e ^{2}} i

\frac{( x ^{2} + 5 x - 36 ) ( x - 4 )}{x + 7}

\frac{a ^{3} + b ^{3} + c ^{3} - 27 ab c}{( a + b ) ( b + c ) ( a + c )}

, \frac{3}{2} (x - 2) - \frac{1}{2} (\frac{x}{3} - 1)

e^{(a + 1) t} sin ((b + 2) x) sin^{2} ((c + 3) x)

\frac{( 8 - x ) ^{5}}{40}

What is i(e^{2+3i}-e^{-2-3i}) ?
The solution to i(e^{2+3i}-e^{-2-3i}) is (-sin(3)-e^4sin(3))/(e^2)+i(-cos(3)+e^4cos(3))/(e^2)