Question

If $x+iy=\sqrt{\varphi +i\Psi }$ where $i=\sqrt{-1}$ and $\varphi$ and $\Psi$ are non zero real parameters then $\varphi =$ constant and $\Psi =\text{constant}$, represents two systems of rectangular hyperbola which intersect at an angle of

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{6}$

Given, $x+iy=\sqrt{\varphi +i\psi }z=x+iy\therefore z^2={(x+iy)}^2=\varphi +i\psi \Rightarrow x^2-y^2+2ixy=\varphi +i\psi u(x,y)=x^2-y^{2}andv(x,y)=2xy$

For all points $x^2-y^2=\varphi$ and in the $XY$ plane

$2xy=\psi$

${(u+iv)}^{1/2}=\pm [\sqrt{\frac{u+\sqrt{u^2+v^2}}{2}}+i\frac{v}{|v|}\sqrt{\frac{\sqrt{u^2+v^2}-u}{2}}]\frac{u+\sqrt{u^2+v^2}}{2}=a^2⟺4a^2(a^2-u)=v^2\frac{\sqrt{u^2+v^2}-u}{2}=b^2⟺4b^2(b^2+u)=v^2$

The complex function $\omega =z^2$ maps a semi-infinite strip onto a semi-infinite parabolic wedge

$R{e}^{i\varphi }=\omega \therefore z=r{e}^{i\theta }\Rightarrow R{e}^{i\varphi }=r^2{e}^{2i\theta }\therefore \varphi =2\theta R=r^{2}r=2(1+\cos \theta )=4{\cos }^2\left(\theta /2\right)\varphi =2\theta \times 3\Rightarrow \theta =\frac{\varphi }{6}=\frac{\pi }{6}$