Question

$\alpha +i\beta ={\tan }^{-1}z,\alpha =\frac{\pi }{4}$ then locus of $z$ is

• A. $x^2+y^2=1$
• B. $x^2+y^2+2x-1=0$
• C. $x^2-y^2=1$
• D. $xy=1$

Answer 1

The correct option is A $x^2+y^2=1$

we have
$\tan (\alpha +i\beta )=z$

$\Rightarrow \frac{\sin (\alpha +i\beta )}{\cos (\alpha +i\beta )}=z$

$\Rightarrow \frac{e^{i(\alpha +i\beta )}-e^{-i(\alpha +i\beta )}}{i(e^{i(\alpha +i\beta )}+e^{-i(\alpha +i\beta )})}=Z$

$\Rightarrow \frac{(e^{i\alpha }.e^{-\beta }-e^{-i\alpha }.e^{\beta })}{i(e^{i\alpha }.e^{-\beta }+e^{-i\alpha }.e^{\beta })}=Z$

$\Rightarrow \frac{e^{2i\alpha }-e^{2i\beta }}{(e^{2i\alpha }+e^{2\beta })}=Zi$
Putting $\alpha =\pi /4$

$\frac{i-e^{2\beta }}{i+e^{2\beta }}=Zi$

$\Rightarrow i-e^{2\beta }=-Z+Zi{r}^{2\beta }$
Putting $e^{2\beta }=k$ and $\Rightarrow i-k=-Z+$

$\alpha +i\beta ={\tan }^{-1}(x+iy)$
Taking conjugate

$\Rightarrow \alpha -i\beta ={\tan }^{-1}(x-iy)$
adding ,

$2\alpha ={\tan }^{-1}(x+iy)+{\tan }^{-1}(x-iy)$
putting $\alpha =\pi /4$

$\Rightarrow \pi /2={\tan }^{-1}\frac{(x+iy+x-y)}{1-(x+iy)(x-iy)}$

$\pi /2={\tan }^{-1}\frac{2x}{1-x^2+y^2}$
which is possible

if $\frac{2x}{1-(x^2+y^2)}\to \infty$

$\Rightarrow 1-(x^2+y^2)=0$
or $x^2+y^2=1$