化简 ((z^2+1)^2)/((z-i)^2)

解答

化简

\frac{( z ^{2} + 1 ) ^{2}}{( z - i ) ^{2}}

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

求解步骤

\frac{( z ^{2} + 1 ) ^{2}}{( z - i ) ^{2}}

展开

(z - i)^{2} : z^{2} - 2 z i - 1

= \frac{( z ^{2} + 1 ) ^{2}}{z ^{2} - 2 z i - 1}

将

z^{2} - 2 z i - 1

改写成标准复数形式:

(z^{2} - 1) - 2 z i

= \frac{( z ^{2} + 1 ) ^{2}}{( z ^{2} - 1 ) - 2 z i}

乘以共轭根式

\frac{( z ^{2} - 1 ) + 2 z i}{( z ^{2} - 1 ) + 2 z i}

= \frac{( z ^{2} + 1 ) ^{2} ( ( z ^{2} - 1 ) + 2 z i )}{( ( z ^{2} - 1 ) - 2 z i ) ( ( z ^{2} - 1 ) + 2 z i )}

化简

\frac{( z ^{2} + 1 ) ^{2} ( ( z ^{2} - 1 ) + 2 z i )}{( ( z ^{2} - 1 ) - 2 z i ) ( ( z ^{2} - 1 ) + 2 z i )} : \frac{( z ^{2} + 1 ) ^{2} ( ( z ^{2} - 1 ) + 2 z i )}{z ^{4} + 2 z ^{2} + 1}

= \frac{( z ^{2} + 1 ) ^{2} ( ( z ^{2} - 1 ) + 2 z i )}{z ^{4} + 2 z ^{2} + 1}

化简

\frac{( z ^{2} + 1 ) ^{2} ( ( z ^{2} - 1 ) + 2 z i )}{z ^{4} + 2 z ^{2} + 1} : z^{2} - 1 + 2 z i

= z^{2} - 1 + 2 z i

Rewrite in standard complex form:

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

= (z^{2} - 1) + 2 z i

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