If x + iy = sqrt{phi + ipsi},where i = sqrt{- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $x + iy = \sqrt{\phi + i\psi}$.Squaring both sides,we get $(x + iy)^2 = \phi + i\psi$.$x^2 - y^2 + 2xyi = \phi + i\psi$.Equating real and im

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Given $x + iy = \sqrt{\phi + i\psi}$.Squaring both sides,we get $(x + iy)^2 = \phi + i\psi$.$x^2 - y^2 + 2xyi = \phi + i\psi$.Equating real and imaginary parts,we have $x^2 - y^2 = \phi$ and $2xy = \psi$.These represent two families of rectangular hyperbolas.Let $f(x, y) = x^2 - y^2 - \phi = 0$ and $g(x, y) = 2xy - \psi = 0$.The gradient of the first curve is $
abla f = (2x, -2y)$ and the gradient of the second curve is $
abla g = (2y, 2x)$.The dot product of the gradients is $(2x)(2y) + (-2y)(2x) = 4xy - 4xy = 0$.Since the dot product is $0$,the curves are orthogonal,meaning they intersect at an angle of $\frac{\pi}{2}$.

If $x + iy = \sqrt{\phi + i\psi}$,where $i = \sqrt{-1}$ and $\phi$ and $\psi$ are non-zero real parameters,then $\phi = \text{constant}$ and $\psi = \text{constant}$ represent two systems of rectangular hyperbolas which intersect at an angle of:

Similar Questions

Geometrically,the set $\{z \in \mathbb{C} : |z - 2 - 2i| \leq 1\}$ represents

Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

If $z$ is a complex number satisfying $|z - 3| \leq 5$,then the range of $|z + 3i|$ is (where $i = \sqrt{-1}$).

Let $S = \{z = x + iy : \frac{2z - 3i}{4z + 2i} \text{ is a real number} \}$. Then which of the following is $NOT$ correct?

The locus of a point $z$ satisfying $|z|^2 = \operatorname{Re}(z)$ is a circle with centre

Vedclass Test Series

Exam Paper Generator

Online Exam Module