化简 e^{2(x+iy)}-2e^{x+iy}+1

解答

化简

e^{2 (x + i y)} - 2 e^{x + i y} + 1

(e^{2 x} cos (2 y) - 2 e^{x} cos (y) + 1) + i (e^{2 x} sin (2 y) - 2 e^{x} sin (y))

求解步骤

e^{2 (x + i y)} - 2 e^{x + i y} + 1

使用虚数运算法则:

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

= e^{2 x} (cos (2 y) + i sin (2 y)) - 2 e^{x + i y} + 1

e^{x + i y} = e^{x} (cos (y) + i sin (y))

= e^{2 x} (cos (2 y) + i sin (2 y)) - 2 e^{x} (cos (y) + i sin (y)) + 1

将

e^{2 x} (cos (2 y) + sin (2 y) i) - 2 e^{x} (cos (y) + sin (y) i) + 1

改写成标准复数形式:

(e^{2 x} cos (2 y) - 2 e^{x} cos (y) + 1) + (e^{2 x} sin (2 y) - 2 e^{x} sin (y)) i

= (e^{2 x} cos (2 y) - 2 e^{x} cos (y) + 1) + (e^{2 x} sin (2 y) - 2 e^{x} sin (y)) i

e x p an d \frac{( 2 - x ) ^{2}}{8}

e x p an d \frac{1 - e ^{- s}}{s ( s ^{2} + 4 )} + \frac{1}{s ^{2} + 4}

e x p an d 2 (x^{2} - y^{2}) d x

\frac{d ^{2}}{d x ^{2}} ((y) (4 cos (3 x) + 6 sin (3 x)))

e x p an d \frac{d}{d x} ((7 x^{2} - 5) e^{x}) - 5 x^{2}