Question
\(\int \frac{z^2-4}{z^2+4} dz\) evaluated anticlockwise around the circle |z - i| = 2 , where i = √−1 , is
The correct answer is
-4π
The contour encloses2i (distance1 from i) and excludes−2i (distance3). Residue at2i is[(2i)²−4]/(4i)=2i. The integral is2πi(2i)=−4π.
Was this answer helpful?
Important Questions from Complex Variables
- Let $z = x + iy$ be a complex variable and $\bar{z}$ be its complex conjugate. The equation$z^2 + \bar{z}^2 = 2$ represents a
- Consider the complex function $f(z) = \cos z + e^{z^2}$. The coefficient of $z^5$ in the Taylor series expansion of $f(z)$ about the origin is ________ (rounded off to 1 decimal place).
- Let $f(z) = j \frac{1-z}{1+z}$, where z denotes a complex number and j denotes $\sqrt{-1}$. The inverse function $f^{-1}(z)$ maps the real axis to the ______.
The contour integral $$\oint_C e^{1/z} dz$$ with C as the counter-clockwise unit circle in the z-plane is equal to
Let $$ I = \int_0^1 \frac{1}{1+t} dt + \frac{\pi i}{2} \int_0^1 \frac{1}{1+e^2} dt - i \int_0^1 \frac{1}{1+it} dt, $$ where $ t $ is a real variable and $ i = \sqrt{-1} $. The value of $ I $ is ______________
Need Expert Advice?
