展开 e^{(a+ib)^2}|e^{(a+ib)^2}|

解答

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e^{(a + ib)^{2}} e^{(a + ib)^{2}}

e^{2 a^{2} - 2 b^{2}} cos (2 ab) + i e^{2 a^{2} - 2 b^{2}} sin (2 ab)

求解步骤

e^{(a + ib)^{2}} e^{(a + ib)^{2}}

e^{(a + ib)^{2}} = e^{a^{2} - b^{2}} cos (2 ab) + i e^{a^{2} - b^{2}} sin (2 ab)

= e^{(a + ib)^{2}} (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab))

(a + ib)^{2} = a^{2} + 2 iab - b^{2}

= e^{a^{2} + 2 iab - b^{2}} (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab))

化简

e^{a^{2} + 2 iab - b^{2}} : e^{a^{2} - b^{2}} (cos (2 ab) + i sin (2 ab))

= e^{a^{2} - b^{2}} (cos (2 ab) + i sin (2 ab)) (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab))

∣ a + bi ∣ = (a + bi) (a - bi) = a^{2} + b^{2}

令

a = e^{a^{2} - b^{2}} cos (2 ab), b = e^{a^{2} - b^{2}} sin (2 ab)

= (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab)) (e^{a^{2} - b^{2}} cos (2 ab))^{2} + (e^{a^{2} - b^{2}} sin (2 ab))^{2}

(e^{a^{2} - b^{2}} cos (2 ab))^{2} + (e^{a^{2} - b^{2}} sin (2 ab))^{2} = e^{2 a^{2} - 2 b^{2}} cos^{2} (2 ab) + e^{2 a^{2} - 2 b^{2}} sin^{2} (2 ab)

= (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab)) e^{2 a^{2} - 2 b^{2}} cos^{2} (2 ab) + e^{2 a^{2} - 2 b^{2}} sin^{2} (2 ab)

e^{2 a^{2} - 2 b^{2}} cos^{2} (2 ab) + e^{2 a^{2} - 2 b^{2}} sin^{2} (2 ab) = e^{a^{2} - b^{2}}

= e^{a^{2} - b^{2}} (i e^{a^{2} - b^{2}} sin (2 ab) + e^{a^{2} - b^{2}} cos (2 ab))

= e^{a^{2} - b^{2}} (e^{a^{2} - b^{2}} cos (2 ab) + i e^{a^{2} - b^{2}} sin (2 ab))

使用分配律:

a (b + c) = ab + a c

a = e^{a^{2} - b^{2}}, b = e^{a^{2} - b^{2}} cos (2 ab), c = i e^{a^{2} - b^{2}} sin (2 ab)

= e^{a^{2} - b^{2}} e^{a^{2} - b^{2}} cos (2 ab) + e^{a^{2} - b^{2}} i e^{a^{2} - b^{2}} sin (2 ab)

= e^{a^{2} - b^{2}} e^{a^{2} - b^{2}} cos (2 ab) + i e^{a^{2} - b^{2}} e^{a^{2} - b^{2}} sin (2 ab)

化简

e^{a^{2} - b^{2}} e^{a^{2} - b^{2}} cos (2 ab) + i e^{a^{2} - b^{2}} e^{a^{2} - b^{2}} sin (2 ab) : e^{2 a^{2} - 2 b^{2}} cos (2 ab) + i e^{2 a^{2} - 2 b^{2}} sin (2 ab)

= e^{2 a^{2} - 2 b^{2}} cos (2 ab) + i e^{2 a^{2} - 2 b^{2}} sin (2 ab)

e x p an d \frac{e ^{5 x}}{( e ^{2 x} - 1 ) ^{2}}

e x p an d 36 (e^{- 0.1 t} - 0.1 e^{- 0.1 t} t)

e x p an d x^{2} + (x + 4) e^{- x}

e x p an d \frac{sin ( 2 x )}{2 cos ^{2} ( x ) \cdot ( 1 + cos ( x ) )}

\frac{\partial}{\partial x} ((f (x, y)) (x^{2} - 2 x y))