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1

### AIEEE 2006

If $${z^2} + z + 1 = 0$$, where z is complex number, then value of $${\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2}$$ is
A
18
B
54
C
6
D
12

## Explanation

$${z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,$$ or $$\,\,\,{\omega ^2}$$

So, $$z + {1 \over z} = \omega + {\omega ^2} = - 1$$

$${z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,$$

$${z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2$$

$${z^4} + {1 \over {{z^4}}} = - 1,$$

$${z^5} + {1 \over {{z^5}}} = - 1$$

and $$\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2$$

$$\therefore$$ The given sum $$= 1 + 1 + 4 + 1 + 1 + 4 = 12$$
2

### AIEEE 2005

If $${z_1}$$ and $${z_2}$$ are two non-zero complex numbers such that $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$, then arg $${z_1}$$ - arg $${z_2}$$ is equal to
A
$${\pi \over 2}\,$$
B
$$- \pi$$
C
0
D
$${{ - \pi } \over 2}$$

## Explanation

Given that, $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$

$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.

S if $${z_1}$$ makes an angle $$\theta$$ with x axis then $${z_2}$$ will also make $$\theta$$ angle.

$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta$$ - $$\theta$$ = 0
3

### AIEEE 2005

If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are
A
$$- 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}$$
B
$$- 1, - 1, - 1$$
C
$$- 1,1 - 2\omega ,1 - 2{\omega ^2}$$
D
$$- 1,1 + 2\omega ,1 + 2{\omega ^2}$$

## Explanation

$${\left( {x - 1} \right)^3} + 8 = 0$$

$$\Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$

$$\Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$

or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$
4

### AIEEE 2005

If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on
A
an ellipse
B
a circle
C
a straight line
D
a parabola

## Explanation

Given $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$

$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$

$$\Rightarrow$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$

$$\Rightarrow$$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)

$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and

$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.

According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points. ### Joint Entrance Examination

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