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1

AIEEE 2006

MCQ (Single Correct Answer)
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 $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]

$$ = - i$$
2

AIEEE 2006

MCQ (Single Correct Answer)
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 $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$
3

AIEEE 2005

MCQ (Single Correct Answer)
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
4

AIEEE 2005

MCQ (Single Correct Answer)
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 $$\,\,\,1 - 2{\omega ^2}.$$

Questions Asked from Complex Numbers

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