簡約 (z^2)/((z+2i)^2*(z^2+9))

解

簡約

\frac{z ^{2}}{( z + 2 i ) ^{2} \cdot ( z ^{2} + 9 )}

\frac{z ^{2} ( ( z ^{2} - 4 ) - 4 z i )}{( z ^{4} + 8 z ^{2} + 16 ) ( z ^{2} + 9 )}

解答ステップ

\frac{z ^{2}}{( z + 2 i ) ^{2} ( z ^{2} + 9 )}

拡張

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

= \frac{z ^{2}}{( z ^{2} + 4 z i - 4 ) ( z ^{2} + 9 )}

標準的な複素数形式で

z^{2} + 4 z i - 4

を書き換える:

(z^{2} - 4) + 4 z i

= \frac{z ^{2}}{( ( z ^{2} - 4 ) + 4 z i ) ( z ^{2} + 9 )}

共役で乗じる

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

= \frac{z ^{2} ( ( z ^{2} - 4 ) - 4 z i )}{( ( z ^{2} - 4 ) + 4 z i ) ( z ^{2} + 9 ) ( ( z ^{2} - 4 ) - 4 z i )}

簡素化

\frac{z ^{2} ( ( z ^{2} - 4 ) - 4 z i )}{( ( z ^{2} - 4 ) + 4 z i ) ( z ^{2} + 9 ) ( ( z ^{2} - 4 ) - 4 z i )} : \frac{z ^{2} ( ( z ^{2} - 4 ) - 4 z i )}{( z ^{4} + 8 z ^{2} + 16 ) ( z ^{2} + 9 )}

= \frac{z ^{2} ( ( z ^{2} - 4 ) - 4 z i )}{( z ^{4} + 8 z ^{2} + 16 ) ( z ^{2} + 9 )}