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the laurent series expansion of f z 2 z 1 z 2 vali
Question:
The Laurent series expansion of
f
(
z
)
=
[
2
(
z
−
1
)
(
z
−
2
)
]
f(z)=[\frac{2}{(z-1)(z-2)}]
f
(
z
)
=
[
(
z
−
1
)
(
z
−
2
)
2
]
valid for
∣
z
−
1
∣
>
1
|z-1|>1
∣
z
−
1∣
>
1
CUET (PG) - 2023
CUET (PG)
Updated On:
Mar 12, 2025
1
(
z
−
1
)
+
2
(
z
−
1
)
3
+
3
(
z
−
1
)
5
+
.
.
.
.
.
\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^5}+.....
(
z
−
1
)
1
+
(
z
−
1
)
3
2
+
(
z
−
1
)
5
3
+
.....
1
(
z
−
1
)
+
2
(
z
−
1
)
3
+
3
(
z
−
1
)
3
+
.
.
.
.
.
\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....
(
z
−
1
)
1
+
(
z
−
1
)
3
2
+
(
z
−
1
)
3
3
+
.....
1
(
z
−
2
)
+
2
(
z
−
2
)
3
+
3
(
z
−
2
)
3
+
.
.
.
.
.
\frac{1}{(z-2)}+\frac{2}{(z-2)^3}+\frac{3}{(z-2)^3}+.....
(
z
−
2
)
1
+
(
z
−
2
)
3
2
+
(
z
−
2
)
3
3
+
.....
1
(
z
−
1
)
+
2
(
z
−
2
)
3
+
3
(
z
−
1
)
3
+
4
(
z
−
2
)
4
+
.
.
.
.
.
\frac{1}{(z-1)}+\frac{2}{(z-2)^3}+\frac{3}{(z-1)^3}+\frac{4}{(z-2)^4}+.....
(
z
−
1
)
1
+
(
z
−
2
)
3
2
+
(
z
−
1
)
3
3
+
(
z
−
2
)
4
4
+
.....
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The Correct Option is
B
Solution and Explanation
The correct option is (B):
1
(
z
−
1
)
+
2
(
z
−
1
)
3
+
3
(
z
−
1
)
3
+
.
.
.
.
.
\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....
(
z
−
1
)
1
+
(
z
−
1
)
3
2
+
(
z
−
1
)
3
3
+
.....
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