Partial fraction of integrand

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  • #1
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I need to find the partial fraction expansion of the integrand z/[(z-2i)(z+i)]
Just doing 1/(z-2i) + 1/(z+i) results in (2z-i)/(z-2i)(z+i).
It seems easy, but I can't figure out what to multiply by to get the correct numerator.
 

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  • #2
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Suppose z/((z - 2i)(z + i)) = A/(z - 2i) + B/(z + i) for all z, where A and B are some real numbers. Multiply both sides with the greatest common denominator to get:

z = (z + i)A + (z - 2i)B.

Then simplify and compare coefficients on both sides...
 
  • #3
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Alright then, so z = (z + i)A + (z - 2i)B, then expanding gives
z = z(A + B) + i(A - 2B)
Since the LHS has no i, then A - 2B = 0, and likewise, A + B = 1
But...this is going nowhere. Where am I slipping up?
 
  • #4
shmoe
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Hi, what do you get for A and B when you solve those equations? They should work.
 
  • #5
dextercioby
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redshift said:
Alright then, so z = (z + i)A + (z - 2i)B, then expanding gives
z = z(A + B) + i(A - 2B)
Since the LHS has no i, then A - 2B = 0, and likewise, A + B = 1
But...this is going nowhere. Where am I slipping up?

What do you mean it's going nowhere? A=2/3;B=1/3.What's wronh with those numbers??
 
  • #6
HallsofIvy
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C'mon now, if you can do integrals of complex numbers, you surely aren't going to let basic algebra stop you!

A- 2B= 0 and A+B= 1. Subtract the second equation from the first to get -3B= -1 or B= 1/3. From that, A- 2(1/3)= 0 gives A= 2/3.

An even simpler way is this: write (as Muzza said)
"z/((z - 2i)(z + i)) = A/(z - 2i) + B/(z + i) for all z, where A and B are some real numbers. Multiply both sides with the greatest common denominator to get:

z = (z + i)A + (z - 2i)B."

Now let z= -i so that (z+i)= 0 and solve for B.

Then let z= 2i so that (z- 2i)= 0 and solve for A.
 

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