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

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(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

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$\frac{\pi}{2}$

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(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:

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