展开 (z^2+i)/((z^3-i)^2)

解答

展开

\frac{z ^{2} + i}{( z ^{3} - i ) ^{2}}

\frac{- z ^{2} + z ^{8} - 2 z ^{3}}{1 + z ^{12} + 2 z ^{6}} + i \frac{- 1 + z ^{6} + 2 z ^{5}}{1 + z ^{12} + 2 z ^{6}}

求解步骤

\frac{z ^{2} + i}{( z ^{3} - i ) ^{2}}

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

= \frac{z ^{2} + i}{z ^{6} - 2 i z ^{3} - 1}

\frac{z ^{2} + i}{z ^{6} - 2 i z ^{3} - 1}

有理化:

\frac{- 2 z ^{3} + z ^{2} ( z ^{6} - 1 ) + i ( z ^{6} + 2 z ^{5} - 1 )}{z ^{12} + 2 z ^{6} + 1}

= \frac{z ^{2} ( z ^{6} - 1 ) - 2 z ^{3} + i ( z ^{6} + 2 z ^{5} - 1 )}{z ^{12} + 2 z ^{6} + 1}

将

\frac{z ^{2} ( z ^{6} - 1 ) - 2 z ^{3} + i ( 2 z ^{5} + z ^{6} - 1 )}{z ^{12} + 2 z ^{6} + 1}

改写成标准复数形式:

\frac{z ^{8} - z ^{2} - 2 z ^{3}}{z ^{12} + 2 z ^{6} + 1} + \frac{2 z ^{5} + z ^{6} - 1}{z ^{12} + 2 z ^{6} + 1} i

= \frac{z ^{8} - z ^{2} - 2 z ^{3}}{z ^{12} + 2 z ^{6} + 1} + \frac{2 z ^{5} + z ^{6} - 1}{z ^{12} + 2 z ^{6} + 1} i

e x p an d (cos (n x) (x + π)) d x

e x p an d \frac{132}{4}

e x p an d 0.5 (3 x - 2 y + 4 z = 11)

e x p an d - 2 (x + 3) (x - 3) + 3 (x^{2} = 2 x + 1)

e x p an d 7 x \cdot 3 y