Evaluate $$ I=\int_C \frac{dz}{z(z-1)(z-2)}$$ where $C = \{z \in \mathbb C: |z| = r\}, r \gt 0$.

I have splitted the integrals using partial fractions and applied Cauchys integral formula and found that

if $r \in (2,3)$ then $I = -2\pi i$

if $r \in (0,1)$ then $I = \pi i$

if $r \in (1,2)$ then $I = -\pi i$

if $r \gt 3$ then $I = -2\pi i$.

But these answers are not in the options..Did i commit any mistake? please help me to find..


the residues at $z=0$ and $z=2$ are each $\pi i$, whilst that at $z=1$ is $-2\pi i$


the simple denominator means that the constants in the partial fraction expansion are easily obtained by inspection using the cover-up technique, which here gives: $$ \frac1{(z-a)(z-b)(z-c)}= \frac{\frac1{(a-b)(a-c)}}{z-a}+\frac{\frac1{(b-a)(b-c)}}{z-b}+\frac{\frac1{(c-a)(c-b)}}{z-c} $$

  • $\begingroup$ thanks for the answer, i just made a mistake and thanks for the shortcut method $\endgroup$ – Sam Christopher May 16 '15 at 6:08

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