What is 1/((z-i)^2) ?

The solution to 1/((z-i)^2) is (z^2-1)/(z^4+2z^2+1)+(2z)/(z^4+2z^2+1)i

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simplify 1/((z-i)^2)

simplify

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

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

Solution steps

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

Expand

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

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

Rewrite

z^{2} - 2 z i - 1

in standard complex form:

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

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

Multiply by the conjugate

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

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

Simplify

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

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

Rewrite in standard complex form:

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

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

3 (\frac{( - 1 ) ^{k - 1} + 1}{2} + 3^{k - 1})

\frac{a + b}{( ( d + e ) \cdot ( a - c ) )}

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

2 cos (t) (3 - 7 tan (t))

\frac{20}{( x - 1 ) ( x + 1 )}

What is 1/((z-i)^2) ?
The solution to 1/((z-i)^2) is (z^2-1)/(z^4+2z^2+1)+(2z)/(z^4+2z^2+1)i