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I do not understand why.. 1/z(z-1)^2 = 1/z (1-z)^-2
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\[\frac{ 1 }{z(z-1)^2 } = \frac{ 1 }{z }(1-z)^-2\]
are you sure you not missing a minus sign?
nope double and triple checked.
that's squared ...
what is squared?
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Maybe you need something along the lines of... \[\Large a^2=(-a)^2\]:D
oh i get what is squared.. the -1 is taken out of the brackets but it has to be squared so becomes 1.. they have done that so it is in the form of the basic Taylor series..
yes terenz...
Then that's settled? Unless I'm missing something, and I highly suspect that to be the case :D
no I think i get it. I was simply over thinking or under thinking.. but now it all makes sense in context to what I am being asked. Thanks
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