What is 1/(w(w+i)^2) ?

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

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simplify 1/(w(w+i)^2)

simplify

\frac{1}{w ( w + i ) ^{2}}

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

Solution steps

\frac{1}{w ( w + i ) ^{2}}

Expand

(w + i)^{2} : w^{2} + 2 w i - 1

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

Rewrite

w^{2} + 2 w i - 1

in standard complex form:

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

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

Multiply by the conjugate

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

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

Simplify

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

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

Rewrite

w^{2} - 1 - 2 w i

in standard complex form:

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

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

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

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

Cancel

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

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

(z + Δ z)^{2}

- \frac{7 ( 3 + 3 )}{6}

\frac{d}{d v} ((p) (p + \frac{2.4}{v ^{2}}) (v - 8.3))

\frac{d ^{3}}{d t ^{3}} ((s) ((3 + 2 t)^{3}))

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