- #1
Noone1982
- 83
- 0
Say we have the function:
[tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}[/tex]
Using partial fractions,
[tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}\; =\; \frac{1}{z-2}\; -\; \frac{1}{z\; -\; 1}[/tex]
My question comes in on why and how these equations are manipulted for different regions.
Now for a) region |z| < 1
[tex] \frac{1}{z-1}\; =\; -\frac{1}{1-z}\; =\; -\sum_{j=0}^{\infty }{z^{j}}\; [/tex]
But for region 1 < |z| < 2
[tex]\frac{1}{z-1}\; =\; \frac{1}{z}\frac{1}{1-\frac{1}{z}}\; =\; -\frac{1}{z}\sum_{j=0}^{\infty }{\frac{1}{z^{j}}\; =\; }\sum_{j=0}^{\infty }{\frac{1}{z^{j+1}}}[/tex]
I have no idea how or why they are being manipulated for different regions. My book assumes me to be brilliant I suppose?
[tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}[/tex]
Using partial fractions,
[tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}\; =\; \frac{1}{z-2}\; -\; \frac{1}{z\; -\; 1}[/tex]
My question comes in on why and how these equations are manipulted for different regions.
Now for a) region |z| < 1
[tex] \frac{1}{z-1}\; =\; -\frac{1}{1-z}\; =\; -\sum_{j=0}^{\infty }{z^{j}}\; [/tex]
But for region 1 < |z| < 2
[tex]\frac{1}{z-1}\; =\; \frac{1}{z}\frac{1}{1-\frac{1}{z}}\; =\; -\frac{1}{z}\sum_{j=0}^{\infty }{\frac{1}{z^{j}}\; =\; }\sum_{j=0}^{\infty }{\frac{1}{z^{j+1}}}[/tex]
I have no idea how or why they are being manipulated for different regions. My book assumes me to be brilliant I suppose?