anonymous
  • anonymous
integral of z^2 / (z-1)3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ash2326
  • ash2326
\[\int \frac{z^2}{(z-1)^3} dz\] Substitute z=u+1 so \[dz=du\] We have now \[\int \frac{(u+1)^2}{(u)^3} dz\] Let's expand the square term \[\int \frac{(u^2+1+2u)}{(u)^3} dz\] we get \[\int \frac{1}{u}+ \frac{1}{u^3}+\frac{2}{u^2} du\] Can you do it now?
anonymous
  • anonymous
but there is another way which i know and that's in which we do something like in numerator we place constants like A B C......
ash2326
  • ash2326
That's the method of Partial fractions. Did you try it?

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anonymous
  • anonymous
the one which you have mentioned i dont used to solve partial fraction question...i used to take that method which i m telling you
anonymous
  • anonymous
z^2= A/(z-1)^2 + B(z-1)
anonymous
  • anonymous
this method will work, but lord is it a pain
anonymous
  • anonymous
you will need \[\frac{z^2}{(z-1)^3}=\frac{a}{z-1}+\frac{b}{(z-1)^2}+\frac{c}{(z-1)^3}\] to do this
anonymous
  • anonymous
\[\frac{z^2}{(z-1)^3}=\frac{1}{z-1}+\frac{2}{(z-1)^2}+\frac{1}{(z-1)^3} \]

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