- #1

- 53

- 0

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter redshift
- Start date

- #1

- 53

- 0

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.

- #2

- 695

- 0

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

Then simplify and compare coefficients on both sides...

- #3

- 53

- 0

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

Science Advisor

Homework Helper

- 1,992

- 1

Hi, what do you get for A and B when you solve those equations? They should work.

- #5

- 13,162

- 725

redshift said:

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

Science Advisor

Homework Helper

- 41,847

- 966

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.

Share: