Let a = lmleft( {frac{{1 + {z^2}}}{{2iz} | vedclass

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Question

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)}

Vedclass · Accepted Answer

$\left( { - 1,1} \right)$

Vedclass · Answer

Let $z=x+i y \Rightarrow z^{2}=x^{2}-y^{2}+2 i x y$

Now,

$\frac{{1 + {z^2}}}{{2iz}} = $ ${ = \frac{{1 + {x^2} - {y^2} + 2ixy}}{{2i(x + iy)}}}$ ${ = \frac{{\left( {{x^2} - {y^2} + 1} ight) + 2ixy}}{{2ix - 2y}}}$

$ = \frac{{\left( {{x^2} - {y^2} + 1} ight) + 2ixy}}{{ - 2y + 2x}} 	imes \frac{{ - 2y - 2ix}}{{ - 2y - 2ix}}$

$ = \frac{{y\left( {{x^2} + {y^2} - 1} ight) + x\left( {{x^2} + {y^2} + 1} ight)i}}{{2\left( {{x^2} + {y^2}} ight)}}$

$a = \frac{{x\left( {{x^2} + {y^2} + 1} ight)}}{{2\left( {{x^2} + {y^2}} ight)}}$

Since, $|z|\,\, = \,1\, \Rightarrow \,\sqrt {{x^2}\, + \,{y^2}} \, = \,1$

$ \Rightarrow \,{x^2}\, + \,{y^2}\, = \,1$

$	herefore \,\,a\,\, = \frac{{x(1 + 1)}}{{2 	imes 1}} = x$

Also ${	ext{z}} 
e {	ext{1}} \Rightarrow {	ext{x + iy}} 
e {	ext{1}}$

$	herefore A = ( - 1,1)$

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

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