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Let $a = lm\left( {\frac{{1 + {z^2}}}{{2iz}}} \right)$, where $z$ is any non-zero complex number. The set $A = \{ a:\left| z \right| = 1\,and\,z \ne \pm 1\} $ is equal to
$\left( { - 1,1} \right)$
$\left[ { - 1,1} \right]$
$\left[ {0,1} \right)$
$\left( { - 1,0} \right]$
Solution
Let $z=x+i y \Rightarrow z^{2}=x^{2}-y^{2}+2 i x y$
Now,
$\frac{{1 + {z^2}}}{{2iz}} = $ ${ = \frac{{1 + {x^2} – {y^2} + 2ixy}}{{2i(x + iy)}}}$ ${ = \frac{{\left( {{x^2} – {y^2} + 1} \right) + 2ixy}}{{2ix – 2y}}}$
$ = \frac{{\left( {{x^2} – {y^2} + 1} \right) + 2ixy}}{{ – 2y + 2x}} \times \frac{{ – 2y – 2ix}}{{ – 2y – 2ix}}$
$ = \frac{{y\left( {{x^2} + {y^2} – 1} \right) + x\left( {{x^2} + {y^2} + 1} \right)i}}{{2\left( {{x^2} + {y^2}} \right)}}$
$a = \frac{{x\left( {{x^2} + {y^2} + 1} \right)}}{{2\left( {{x^2} + {y^2}} \right)}}$
Since, $|z|\,\, = \,1\, \Rightarrow \,\sqrt {{x^2}\, + \,{y^2}} \, = \,1$
$ \Rightarrow \,{x^2}\, + \,{y^2}\, = \,1$
$\therefore \,\,a\,\, = \frac{{x(1 + 1)}}{{2 \times 1}} = x$
Also ${\text{z}} \ne {\text{1}} \Rightarrow {\text{x + iy}} \ne {\text{1}}$
$\therefore A = ( – 1,1)$
