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1

### 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.
2

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

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

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

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

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

= $$-2$$
3

### 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) = $${{3\pi } \over 4}$$
4

### 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.

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