Equation Transformation from Complex to Cartesian
We are given the equation involving a complex variable $z = x + iy$ and its conjugate $\bar{z} = x - iy$: $z^2 + \bar{z}^2 = 2$
Deriving the Cartesian Equation
- Substitute $z$ and $\bar{z}$ into the equation:
$(x + iy)^2 + (x - iy)^2 = 2$
- Expand the squared terms:
$(x^2 + 2ixy + (iy)^2) + (x^2 - 2ixy + (iy)^2) = 2$
$(x^2 + 2ixy - y^2) + (x^2 - 2ixy - y^2) = 2$
- Combine like terms. Notice that the imaginary parts ($2ixy$ and $-2ixy$) cancel out:
$x^2 - y^2 + x^2 - y^2 = 2$
$2x^2 - 2y^2 = 2$
- Simplify the equation by dividing by 2:
$x^2 - y^2 = 1$
Identifying the Geometric Shape
The equation $x^2 - y^2 = 1$ is the standard form of a hyperbola centered at the origin with its transverse axis along the x-axis.
Therefore, the equation $z^2 + \bar{z}^2 = 2$ represents a hyperbola.
Important Questions from Complex Variables
- 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 ______________
- Given $f(z) = \frac{1}{z+1} - \frac{2}{z+3}$. If $C$ is a counterclockwise path in the $z$-plane such that $|z+1|=1$, the value of $\frac{1}{2\pi j}\oint_C f(z)dz$ is
