 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

If $$z = x - iy$$ and $${z^{{1 \over 3}}} = p + iq$$, then

$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to
A
- 2
B
- 1
C
2
D
1

## Explanation

Given $${z^{{1 \over 3}}} = p + iq$$

$$\Rightarrow$$ z = (p + iq)3

= p3 + (iq)3 +3p(iq)(p + iq)

= p3 - iq3 +3ip2q - 3pq2

= p(p2 - 3q2) - iq(q2 - 3p2)

Given that $$z = x - iy$$

$$\therefore$$ $$x - iy$$ = p(p2 - 3q2) - iq(q2 - 3p2)

By comparing both sides we get,

$${x \over p} = {p^2} - 3{q^2}$$ and $${y \over q} = {q^2} - 3{p^2}$$

$$\therefore$$ $${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$

= $${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$$

= $$-2$$
2

### AIEEE 2004

Let z and w be complex numbers such that $$\overline z + i\overline w = 0$$ and arg zw = $$\pi$$. Then arg z equals
A
$${{5\pi } \over 4}$$
B
$${{\pi } \over 2}$$
C
$${{3\pi } \over 4}$$
D
$${{\pi } \over 4}$$

## Explanation

Given $$\overline z + i\overline w = 0$$

$$\Rightarrow \overline z = - i\overline w$$

$$\Rightarrow \overline{\overline z} = - \overline {i\overline w }$$

$$\Rightarrow \overline{\overline z} = - \overline i \overline{\overline w}$$

$$\Rightarrow z = - \overline i w$$

$$\Rightarrow z = - \left( { - i} \right)w$$

$$\Rightarrow z = iw$$

Now given that Arg(zw) = $$\pi$$

$$\Rightarrow$$ Arg(z$$\times$$$${z \over i}$$) = $$\pi$$

$$\Rightarrow$$ Arg(z2) - Arg(i) = $$\pi$$

$$\Rightarrow$$ 2Arg(z) - $${\pi \over 2}$$ = $$\pi$$

[ $$i$$ complex number represent (0, 1) point on imaginary axis and Arg($$i$$) means the angle made by the point (0, 1) with real axis which is $${\pi \over 2}$$]

$$\Rightarrow$$ 2Arg(z) = $${{3\pi } \over 2}$$

$$\Rightarrow$$ Arg(z) = $${{3\pi } \over 4}$$
3

### AIEEE 2003

If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then
A
x = 2n + 1, where n is any positive integer
B
x = 4n , where n is any positive integer
C
x = 2n, where n is any positive integer
D
x = 4n + 1, where n is any positive integer.

## Explanation

$${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$

$$\Rightarrow$$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$

$$\Rightarrow$$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$

$$\Rightarrow$$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$

$$\Rightarrow$$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$

$$\Rightarrow {\left( i \right)^x} = 1$$

We know $$i = \sqrt { - 1}$$

$$\therefore$$ $${i^2} = - 1$$

$$\Rightarrow$$ $${i^3} = - 1 \times i = - i$$

$$\Rightarrow$$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$

So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1

$$\therefore$$ $${\left( i \right)^x} = 1$$ $$= {\left( i \right)^{4n}}$$ where n is a positive integer.
4

### AIEEE 2003

Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then
A
$${a^2} = 4b$$
B
$${a^2} = b$$
C
$${a^2} = 2b$$
D
$${a^2} = 3b$$

## Explanation

$${Z^2} + aZ + b = 0$$
and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,
There are three complex numbers:
1. Origin (0)
2. $${Z_1}$$
3. $${Z_2}$$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,
$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$\Rightarrow$$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$\Rightarrow$$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$\Rightarrow$$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$\Rightarrow$$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$\Rightarrow$$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]

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