What is 3/((z-i))-6/((z-i)^2) ?

The solution to 3/((z-i))-6/((z-i)^2) is (3z^3+3z-6z^2+6)/((z^2+1)^2)+(3+3z^2-12z)/((z^2+1)^2)i

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simplify 3/((z-i))-6/((z-i)^2)

simplify

\frac{3}{( z - i )} - \frac{6}{( z - i ) ^{2}}

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

Solution steps

\frac{3}{z - i} - \frac{6}{( z - i ) ^{2}}

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

\frac{6}{( z - i ) ^{2}} = \frac{6 z ^{2} - 6 + 12 z i}{z ^{4} + 2 z ^{2} + 1}

= \frac{3 z + 3 i}{z ^{2} + 1} - \frac{6 z ^{2} - 6 + 12 z i}{z ^{4} + 2 z ^{2} + 1}

Join

\frac{3 z + 3 i}{z ^{2} + 1} - \frac{6 z ^{2} - 6 + 12 z i}{z ^{4} + 2 z ^{2} + 1} : \frac{3 z ^{3} + 3 z + 3 i + 3 z ^{2} i - 6 z ^{2} + 6 - 12 i z}{( z ^{2} + 1 ) ^{2}}

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

Rewrite in standard complex form:

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

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

(a - b)^{3} - (b - a)^{3}

(3 m^{- 2} n^{4})^{- 2}

(1.5 a + 1.3) (0.4 a^{2} - 2 a + 3)

\frac{( x - 4 )}{( x - 4 ^{2} ) \cdot ( x + 4 )}

(y^{2} - y + 3) (y^{2} + y - 3)

What is 3/((z-i))-6/((z-i)^2) ?
The solution to 3/((z-i))-6/((z-i)^2) is (3z^3+3z-6z^2+6)/((z^2+1)^2)+(3+3z^2-12z)/((z^2+1)^2)i