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1

### 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 $$\,\,\,1 - 2{\omega ^2}.$$
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 2004

If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on
A
an ellipse
B
the imaginary axis
C
a circle
D
the real axis

## Explanation

Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$,

By squaring both sides we get,

$${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$\Rightarrow$$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z$$ ]

$$\Rightarrow$$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$

$$\Rightarrow$$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$\Rightarrow$$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$\Rightarrow$$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$\Rightarrow$$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$\Rightarrow$$ $${z + \overline z }$$ = 0

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

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$\Rightarrow$$ x + iy = - x + iy

$$\Rightarrow$$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.
4

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

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

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

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

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

= $$-2$$

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