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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