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1

### AIEEE 2007

If $$\,\left| {z + 4} \right|\,\, \le \,\,3\,$$, then the maximum value of $$\left| {z + 1} \right|$$ is
A
6
B
0
C
4
D
10

## Explanation

$$z$$ lies on or inside the circle with center $$(-4,0)$$ and radius $$3$$ units.

From the Argand diagram maximum value of $$\left| {z + 1} \right|$$ is $$6$$
2

### AIEEE 2006

The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)}$$ is
A
i
B
1
C
- 1
D
- i

## Explanation

$$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)}$$

$$= i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)}$$

$$= i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}$$

$$= i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$$

$$= i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$= i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$= i \times 0 - i$$

[as $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]

$$= - i$$
3

### 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 $$= 1 + 1 + 4 + 1 + 1 + 4 = 12$$
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.

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