Could someone explain why $z = \frac{a}{b}$ is a singularity of $ \frac{1}{z|az-b|^2} $?
Am I mistaken something?
Asked
Active
Viewed 40 times
4
-
1Do you mean $z=\frac{b}a$ is a singularity? @Ignite – Olivier Oloa Dec 20 '16 at 11:02
-
Did you mean $z=b/a$? If so, because the expression is not defined when the denominator is 0. – Rumplestillskin Dec 20 '16 at 11:03
-
1No, it is $\frac{a}{b}$ – IgNite Dec 20 '16 at 11:03
1 Answers
4
Assume $a\neq0,\, b \neq0$. We have $$ a z -b=0 \Leftrightarrow z=\frac{b}{a}, $$ then $\displaystyle z=\frac{a}{b}$ is not a singularity of the denominator unless $$ a \cdot\frac{a}{b} -b=0 \Leftrightarrow a^2=b^2. $$
Olivier Oloa
- 120,989
-
2The last line makes a lot of sense $-$ "unless". At a first glance $\frac{a}{b}$ doesn't seem to be the zero of $z|az-b|^2$, but after take a careful look, it is. – IgNite Dec 20 '16 at 11:14