How do you write the partial fraction decomposition of the rational expression #z/(z^4-1) #?

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Answer 1 · 2015-12-13T19:39:36

Answer is given as follows:

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Answer 2 · 2015-12-13T23:02:22

#z/(z^4-1)=1/(4(z+1))+1/(4(z-1))-z/(2(z^2+1))#

The expression equals #z/((z+1)(z-1)(z^2+1)#.

#z/((z+1)(z-1)(z^2+1))=A/(z+1)+B/(z-1)+(Cz+D)/(z^2+1)#

#z=A(z-1)(z^2+1)+B(z+1)(z^2+1)+(Cz+D)(z^2-1)#

IF #z=1#:

#1=4B#
#B=1/4#

IF #z=-1#:

#-1=-4A#
#A=1/4#

Plug in two values for #z#:

IF #z=2#:

#-1=2C+D#

IF #z=3#:

#-3/2=3C+D#

Solve the system to find that #C=-1/2# and #D=0#.

Thus:

#z/(z^4-1)=1/(4(z+1))+1/(4(z-1))-z/(2(z^2+1))#