A simple closed path 𝐶 in the complex plane is shown in the figure if
\[
\oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A
\]
where 𝑖=√−1, then the value of 𝐴 is ______ (rounded off to two decimal places).
\[
\oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A
\]
\[
\oint_C \frac{2^z}{z^2-1}\,dz
=
\oint_C
\frac{2^z}{(z+1)(z-1)}\,dz
\]
Residue at \(z=1\):
\[
\operatorname{Res}_{z=1}
\frac{2^z}{(z+1)(z-1)}
=
\lim_{z\to1}
(z-1)
\frac{2^z}{(z+1)(z-1)}
=
\frac{2}{2}
=1
\]
\[
\oint_C
\frac{2^z}{z^2-1}\,dz
=
2\pi i
\]
Residue at \(z=-1\):
\[
\operatorname{Res}_{z=-1}
\frac{2^z}{(z+1)(z-1)}
=
\lim_{z\to-1}
(z+1)
\frac{2^z}{(z+1)(z-1)}
=
\frac{2^{-1}}{-2}
=
-\frac14
\]
\[
\oint_C
\frac{2^z}{z^2-1}\,dz
=
-\frac14(2\pi i)
=
-\frac12\pi i
\]
Given:
\[
\oint_C
\frac{2^z}{z^2-1}\,dz
=
-i\pi A
\]
Therefore,
\[
-i\pi A
=
-\frac12\pi i
\]
\[
A=\frac12=0.50
\]
Note: Here only \(z=-1\) lies inside the contour.