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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
A value of $$\theta \,$$ for which $${{2 + 3i\sin \theta \,} \over {1 - 2i\,\,\sin \,\theta \,}}$$ is purely imaginary, is :
A
$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 4}} \right)$$
B
$${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)\,$$
C
$${\pi \over 3}$$
D
$${\pi \over 6}$$

Explanation

Rationalizing the given expression

$${{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.

$$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$$

$$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$$

$$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$$
2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
A complex number z is said to be unimodular if $$\,\left| z \right| = 1$$. Suppose $${z_1}$$ and $${z_2}$$ are complex numbers such that $${{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}$$ is unimodular and $${z_2}$$ is not unimodular. Then the point $${z_1}$$ lies on a :
A
circle of radius 2.
B
circle of radius $${\sqrt 2 }$$.
C
straight line parallel to x-axis
D
straight line parallel to y-axis.

Explanation

$$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$$

$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$$

$$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$$

$$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$$

$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$$

$$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$$

$$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$$

As $$\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$$ $$\therefore$$ $$\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$$

$$ \Rightarrow \,$$ Point $$\,{z_1}\,$$ lies on circle of radius $$2.$$
3

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
If z is a complex number such that $$\,\left| z \right| \ge 2\,$$, then the minimum value of $$\,\,\left| {z + {1 \over 2}} \right|$$ :
A
is strictly greater that $${{5 \over 2}}$$
B
is strictly greater that $${{3 \over 2}}$$ but less than $${{5 \over 2}}$$
C
is equal to $${{5 \over 2}}$$
D
lie in the interval (1, 2)

Explanation

We know minimum value of

$$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$$

Thus minimum value of

$$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$$ is $$\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$$

Since, $$\,\,\,\left| Z \right| \ge 2$$

$$\therefore$$ $$\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$$

$$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$$
4

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
If z is a complex number of unit modulus and argument $$\theta $$, then arg $$\left( {{{1 + z} \over {1 + \overline z }}} \right)$$ equals :
A
$$ - \theta \,\,$$
B
$${\pi \over 2} - \theta \,$$
C
$$\theta \,$$
D
$$\,\pi - \theta \,\,$$

Explanation

Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $$

As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$

$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$

$$ = \arg \left( z \right) = \theta .$$

Questions Asked from Complex Numbers

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