Let a = lmleft( {frac{{1 + {z^2}}}{{2iz}}} right) , where z is any non-zero complex numbe

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4-1.Complex numbers

hard

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

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

$\left[ { - 1,1} \right]$

$\left[ {0,1} \right)$

$\left( { - 1,0} \right]$

(JEE MAIN-2013)

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 ${\text{z}} \ne {\text{1}} \Rightarrow {\text{x + iy}} \ne {\text{1}}$

$\therefore A = ( – 1,1)$

Standard 11

Mathematics

If $z_1 = 1+2i$ and $z_2 = 3+5i$ , then ${\mathop{\rm Re}\nolimits} \,\left( {\frac{{{{\overline Z }_2}{Z_1}}}{{{Z_2}}}} \right) = $

normal

For any complex number $w = c + id$, let $\arg ( w ) \in(-\pi, \pi]$, where $i =\sqrt{-1}$. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers $z=x+$ iy satisfying arg $\left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$, the ordered pair $( x , y )$ lies on the circle

$x^2+y^2+5 x-3 y+4=0 .$

Then which of the following statements is (are) TRUE?

$(A)$ $\alpha=-1$ $(B)$ $\alpha \beta=4$ $(C)$ $\alpha \beta=-4$ $(D)$ $\beta=4$

medium

(IIT-2021)

Let $z$ be complex number satisfying $|z|^3+2 z^2+4 z-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.

Match each entry in List-$I$ to the correct entries in List-$II$.

List-$I$ List-$II$

($P$) $|z|^2$ is equal to ($1$) $12$

($Q$) $|z-\bar{z}|^2$ is equal to ($2$) $4$

($R$) $|z|^2+|z+\bar{z}|^2$ is equal to ($3$) $8$

($S$) $|z+1|^2$ is equal to ($4$) $10$

($5$) $7$

The correct option is:

normal

(IIT-2023)

If $z$ is a complex number such that $|z – \bar{z}| = 2$ and $|z + \bar{z}| = 4 $, then which of the following is always incorrect –

normal

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation

$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is. . . . . .

normal

(IIT-2022)

List-$I$	List-$II$
($P$) $\|z\|^2$ is equal to	($1$) $12$
($Q$) $\|z-\bar{z}\|^2$ is equal to	($2$) $4$
($R$) $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	($3$) $8$
($S$) $\|z+1\|^2$ is equal to	($4$) $10$
	($5$) $7$

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

Solution

Similar Questions

If $z_1 = 1+2i$ and $z_2 = 3+5i$ , then ${\mathop{\rm Re}\nolimits} \,\left( {\frac{{{{\overline Z }_2}{Z_1}}}{{{Z_2}}}} \right) = $

If $z$ is a complex number such that $|z – \bar{z}| = 2$ and $|z + \bar{z}| = 4 $, then which of the following is always incorrect –

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is. . . . . .

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Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation

$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is. . . . . .