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1

### JEE Main 2019 (Online) 11th January Evening Slot

Let z be a complex number such that |z| + z = 3 + i (where i = $$\sqrt { - 1}$$). Then |z| is equal to
A
$${{\sqrt {34} } \over 3}$$
B
$${5 \over 3}$$
C
$${5 \over 4}$$
D
$${{\sqrt {41} } \over 4}$$

## Explanation

$$\left| z \right| + z = 3 + i$$

$$z = 3 - \left| z \right| + i$$

Let  $$3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$$

$$\Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1}$$

$$\Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}$$

$$\Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}$$
2

### JEE Main 2019 (Online) 11th January Morning Slot

Let $${\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,$$ where x and y are real numbers, then y $$-$$ x equals :
A
$$-$$ 85
B
85
C
$$-$$ 91
D
91

## Explanation

$${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$$

$$= {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$$

Hence, $$y - x = 198 - 107 = 91$$
3

### JEE Main 2019 (Online) 10th January Evening Slot

Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then -
A
R(z) = $$-$$ 3
B
R(z) < 0 and I(z) > 0
C
I(z) = 0
D
R(z) > 0 and I(z) > 0

## Explanation

$$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$$

$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$

$$= {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$

$$= \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$$

$$= 2\cos {{5\pi } \over 6} < 0$$

$${\rm I}(z) = 0$$ and $${\mathop{\rm Re}\nolimits} (z) < 0$$
4

### JEE Main 2019 (Online) 10th January Morning Slot

Let z1 and z2 be any two non-zero complex numbers such that   $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$  If  $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$  then -
A
$${\rm I}m\left( z \right) = 0$$
B
$$\left| z \right| = \sqrt {{17 \over 2}}$$
C
$$\left| z \right| =$$ $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }$$
D
Re(z) $$=$$ 0

## Explanation

Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$

$$\Rightarrow$$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$\Rightarrow$$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta$$ + i sin$$\theta$$)

= 2(cos$$\theta$$ + i sin$$\theta$$)

$$\therefore$$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$
Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta$$ + i sin$$\theta$$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta }$$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }$$

z is neither purely real nor purely imaginary and |z| depends on $$\theta$$.

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