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1

### JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
If $${{z - \alpha } \over {z + \alpha }}\left( {\alpha \in R} \right)$$ is a purely imaginary number and | z | = 2, then a value of $$\alpha$$ is :
A
$${1 \over 2}$$
B
$$\sqrt 2$$
C
2
D
1

## Explanation

$${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$$

$$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$$

$${\left| z \right|^2} = {\alpha ^2},$$  $$a = \pm 2$$
2

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

MCQ (Single Correct Answer)
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  $$3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$$

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

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

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

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

MCQ (Single Correct Answer)
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, $$y - x = 198 - 107 = 91$$
4

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

MCQ (Single Correct Answer)
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 $${\mathop{\rm Re}\nolimits} (z) < 0$$

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