Laurent Series Harder Example

[@sudoLife]

Thanks, very helpful! One thing to note is, instead of doing guesswork to derive a needed expression we can just write 1/(z + z^2) = X * 1 / (1 - 1/(z+1)) and find X. It seems faster this way.

[@7phanthony8]

Nice explanation, thank you!

[@lucassignorini2376]

Thank you , from Argentina :)

[@TheMathSorcerer]

[@SirGumbi]

PERFECT video. Thank you!!!

[@vakhariyajay2224]

Thank you very much. 👍👍🙏🙏🔝🔝👌👌

[@TheMathSorcerer]

Welcome 👍

[@WA-hq6ls]

Very nice Video, straight to the points

[@TheMathSorcerer]

thank you!

[@desmond8745]

Thanks

[@jinuj171]

Can this type of questions be asked in engineering exams , i felt difficult by doing it

[@inasfatiha4778]

Thank uuu so much u saved me

[@TheMathSorcerer]

:)

[@desmond8745]

last minute saviour

[@tnboys9773]

represent the functionz+1/z-1 by irs laurent series in powers of z for the region |z|>1.Answer bro please

[@akshaynayak9171]

can't see what is written

[@DuckStrider]

Ty

[@filipefernandes5721]

So in this case the residue is zero?

[@christianharriviktorreibol5521]

residue is only defined, if we calculate a laurent series around singularities. In this example we calculate the laurant- seies around the circle with radius 1, if i am not mistaken

[@EmperorMussa]

5:02