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1

### JEE Main 2013 (Offline)

If z is a complex number of unit modulus and argument $$\theta$$, then arg $$\left( {{{1 + z} \over {1 + \overline z }}} \right)$$ equals :
A
$$- \theta \,\,$$
B
$${\pi \over 2} - \theta \,$$
C
$$\theta \,$$
D
$$\,\pi - \theta \,\,$$

## Explanation

Given $$\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta$$

As we know, $$\,\,\,\,\overrightarrow z = {1 \over z}$$

$$\therefore$$ $$\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$$

$$= \arg \left( z \right) = \theta .$$
2

### AIEEE 2012

If $$z \ne 1$$ and $$\,{{{z^2}} \over {z - 1}}\,$$ is real, then the point represented by the complex number z lies :
A
either on the real axis or a circle passing through the origin.
B
on a circle with centre at the origin
C
either on real axis or on a circle not passing through the origin.
D
on the imaginary axis.

## Explanation

Let $$z = x + iy$$

$$\therefore$$ $$\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$$

Now $${{{z^2}} \over {z - 1}}$$ is real

$$\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$$

$$\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$$

$$\Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$$

$$\Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0$$

$$\Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0$$

$$\Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$$

$$\therefore$$ $$\,\,\,\,$$ $$z$$ lies either on real axis or on a circle through origin.
3

### AIEEE 2011

If $$\omega ( \ne 1)$$ is a cube root of unity, and $${(1 + \omega )^7} = A + B\omega \,$$. Then $$(A,B)$$ equals
A
(1 ,1)
B
(1, 0)
C
(- 1 ,1)
D
(0 ,1)

## Explanation

$${\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega$$

$$- {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega$$

$$\Rightarrow A = 1,B = 1.$$
4

### AIEEE 2011

Let $$\alpha \,,\beta$$ be real and z be a complex number. If $${z^2} + \alpha z + \beta = 0$$ has two distinct roots on the line Re z = 1, then it is necessary that :
A
$$\beta \, \in ( - 1,0)$$
B
$$\left| {\beta \,} \right| = 1$$
C
$$\beta \, \in (1,\infty )$$
D
$$\beta \, \in (0,1)$$

## Explanation

As real part of roots is $$1$$

Let roots are $$1 + pi,1 + q$$

$$\therefore$$ sum of roots $$= 1 + pi + 1 + qi = - \alpha$$

which is real $$\Rightarrow q = - p\,\,$$

or root are $$1+pi$$ and $$1-pi$$

product of roots $$= 1 + {p^2} = \beta \in \left( {1,\infty } \right)$$

$$p \ne 0$$ as roots are distinct.

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