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
If z is a complex variable, the value of \(\mathop \smallint \limits_5^{3{\rm{i}}} \frac{{{\rm{dz}}}}{{\rm{z}}}\) is
- \(\cos \frac{\pi}{3}+\frac{1}{2} \cos \frac{2 \pi}{3}\) + \(\frac{1}{3} \cos \frac{3 \pi}{3} \ldots \infty\) = will
Imaginary part of \(\cos ^{-1}\left(\frac{3-2 i}{3+2 i}\right)\) = ______
The modulus of 1 + cos α + i sin α 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 \(\frac{1}{2\pi i}\int_c f(z)dz\) is
