IIT Previous Years Solved Examples : Complex Number

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Ques ( JEE Main 2013) : If z is a complex number of unit modulus and argument $\theta$ , then $\arg .\left( {\frac{{1 + z}}{{1 + \overline z }}} \right)$ is equal to

a) - $\theta$ b) $\frac{\pi }{2} - \theta$ c) $\theta$ d) $\pi - \theta$

Solution: Since |z| = 1, $\arg .z = \theta$

Always remember, $z.\overline z = |z{|^2}$

or, $z.\overline z = 1 \Rightarrow \overline z = \frac{1}{z}$ ( as |z| = 1)

Therefore, $\arg .\left( {\frac{{1 + z}}{{1 + \bar z}}} \right) = \arg .\left( {\frac{{1 + z}}{{1 + \frac{1}{z}}}} \right) = \arg .z = \theta$ , Correct option is (c)

Ques: (JEE Advanced 2013) Let $w = \frac{{\sqrt 3 + i}}{2}$ and $P = \{ {w^n}:n = 1,2,3,....\}$ . Further ${H_1} = \{ z \in C:{{\rm Re}\nolimits} z > \frac{1}{2}\}$ and ${H_2} = \{ z \in C:{{\rm Re}\nolimits} z < - \frac{1}{2}\}$ where C is the set of all complex numbers, if ${z_1} \in P\bigcap {{H_1}}$ and ${z_2} \in P\bigcap {{H_2}}$ and O represents the origin, then angle ${z_1}O{z_2}$ is equal to

(a) pi/2 (b) pi/6 (c) 2pi/3 (d) 5pi/6

Solution: Since $w = \frac{{\sqrt 3 + i}}{2}$

or, $\omega = \cos \frac{\pi }{6} + i\sin \frac{\pi }{6}$

and , $P = \{ {w^n}:n = 1,2,3,....\}$

or, $P = \left\{ {\cos \frac{{n\pi }}{6} + i\sin \frac{{n\pi }}{6}:n = 1,2,3....} \right\}$

Therefore, multiple of $\cos \frac{\pi }{6}$ having distinct values are $\frac{{\sqrt 3 }}{2}$ , 1/2 , 0 , -1/2 , $\frac{{ - \sqrt 3 }}{2}$ , -1 , 1
Therefore P has 7 elements

Now, ${H_1} = \{ z \in C:{{\rm Re}\nolimits} z > \frac{1}{2}\}$

Therefore, ${z_1} \in P\bigcap {{H_1}}$

Since, ${{\rm Re}\nolimits} (z) > \frac{1}{2}$

The common elements with P should have ${{\rm Re}\nolimits} (z) = \cos \frac{{n\pi }}{6} > \frac{1}{2}$

Therefore Real(z1) may have the values $\frac{{\sqrt 3 }}{2}$ , 1 i.e. Arg(z1) may have the values pi/6, -pi/6 and 0.
as Principal Argument belongs to (-pi, pi]

And, ${H_2} = \{ z \in C:{{\rm Re}\nolimits} z < - \frac{1}{2}\}$

Therefore, ${z_2} \in P\bigcap {{H_2}}$

The common elements with P should have ${{\rm Re}\nolimits} (z) = \cos \frac{{n\pi }}{6} < - \frac{1}{2}$
Therefore Real(z2) may have the values $\frac{{ - \sqrt 3 }}{2}$ , 1 i.e. Arg(z2) may have the values 5pi/6, -5pi/6 and pi.

From figure it is clear that the angle z1Oz2 may be 2pi/3 ,5pi/6, and pi

So options (c) and (d) are correct.

Ques (IIT 2007): If |z| = 1 and $z \ne \pm 1$ , then all the values of $\frac{z}{{1 - {z^2}}}$ lie on

a) a line not passing through the origin
b) |z| = root 2
c) the x-axis
d) the y-axis

Sol: |z| = 1

Always remember, $z.\bar z = |z{|^2} = 1$ ( as |z| = 1 )

Therefore, $\frac{z}{{1 - {z^2}}} = \frac{z}{{z.\bar z - {z^2}}}$

$= \frac{z}{{z(\bar z - z)}}$

$= \frac{1}{{\bar z - z}}$

Since, $\bar z - z = Purely{{\rm Im}\nolimits}$
Therefore, $\frac{1}{{\bar z - z}}$ is purely imaginary

as if z = x + iy, $\bar z = x - iy$

therefore, $\frac{1}{{\bar z - z}} = \frac{1}{{ - 2iy}}$

or, $\frac{1}{{\bar z - z}} = \frac{i}{{2y}}$ = Purely imaginary

therefore values of $\frac{z}{{1 - {z^2}}}$ lie on y-axis

correct option is (d)

Ques (IIT 2009) Let z = x + iy be a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation $z{\overline z ^3} + \overline z {z^3} = 350$

Solution: Given equation is $z{\overline z ^3} + \overline z {z^3} = 350$

its a four degree equation giving four values.

Always remember, $z.\overline z = |z{|^2}$

Therefore, $z{\overline z ^3} + \overline z {z^3} = 350$

or, $z\overline z {\overline z ^2} + \overline z z{z^2} = 350$

or, $|z{|^2}{\overline z ^2} + |z{|^2}{z^2} = 350$

or, $|z{|^2}({\overline z ^2} + {z^2}) = 350$

or, $({x^2} + {y^2})\{ {(x - iy)^2} + {(x + iy)^2}\} = 350$

or, $({x^2} + {y^2})(2{x^2} - 2{y^2}) = 350$

or, $({x^2} + {y^2})({x^2} - {y^2}) = 175$

Since x and y are integers,

therefore, $({x^2} + {y^2})$ and $({x^2} - {y^2})$ are integers and sum and difference of squares of two numbers.

factor of 175 = 25*7 ,

25 = 4^2 + 3^2
7 = 4^2 - 3^2

Therefore, ${x^2} = 16 \Rightarrow x = \pm 4$
${y^2} = 9 \Rightarrow y = \pm 3$

Therefore four points are (4,3) , (-4,3) , (4,-3) and (-4,-3)

Therefore, area of rectangle = 8 X 6 = 48 square units (option (c) is correct )

Ques: (IIT 2011) If z is any complex number satisfying $|z - 3 - 2i| \le 2$ , then the maximum value of |2z - 6 + 5i| is ........................

Solution: Consider $|z - 3 - 2i| \le 2$

It is the region under the circle having centre (3,2) and radius 2 including the boundary of the circle.( see the graphical representation of complex number - click here )

and we may write |2z - 6 + 5i| = 2|z - 3 + 5/2 i|

Therefore, |z - 3 + 5/2 i| is distance of 'z' from the point (3,-5/2)

Since PA = 6.5

Therefore Max ( |z - 3 + 2.5i| ) = 6.5

Therefore, Max 2|z - 3 + 2.5i| = Max |2z - 6 + 5i| = 2*6.5 = 13 (answer)

Ques (IIT 2003) : If |z| = 1 and $w = \frac{{z - 1}}{{z + 1}}$ ( where $z \ne - 1$ ). then Re (w) is

a) 0
b) $\frac{1}{{|z + 1{|^2}}}$
c) $\left| {\frac{1}{{z + 1}}} \right|\frac{1}{{|z + 1{|^2}}}$
d) $\frac{{\sqrt 2 }}{{|z + 1{|^2}}}$

Solution: Since, $w = \frac{{z - 1}}{{z + 1}}$ ( since $z \ne - 1$ therefore denominator cannot be zero and so well defined )

or, wz + w = z - 1

or, w + 1 = z (1 - w)

or, $z = \frac{{w + 1}}{{1 - w}}$

or, $\left| z \right| = \left| {\frac{{w + 1}}{{1 - w}}} \right|$

or, $1 = \left| {\frac{{w + 1}}{{1 - w}}} \right|$ ( since |z| = 1)

or, |w - 1| = |w + 1|

i.e. distance of w is equal from 1 and -1

therefore w is on the perpendicular bisector of the line joining 1 and -1

therefore w is on the imaginary axis.

or, w is purely real

( see the graphical representation of complex number - click here )

=> Re(w) = 0

Therefore, option (a) is correct

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IIT Previous Years Solved Examples : Complex Number

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