Solved questions by using geometrical representation of complex number

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1. If  |z - i| \le 1 then find the complex no. for which | z - 1 | is least and maximum and find the least and maximum value of | z - 1|

Sol. Since |z - i| \le 1 represent a region under the circle of radius 1 and centre at (0,1). It also includes the perimetre of the circle, | z - 1 | can also equal to 1


We have to find the least and Max value of | z - 1 |, | z - 1 | is the distance between the complex no. 1 and z.

 Since 'z' belongs to the region under the circle.

 so we have to find that point on the region which is at the least distance from the point (1,0) i.e. complex no. 1 + 0i

 By, geometry AP is the least distance from the region of disk and BP is the maximum distance.

 Therefore,|z - 1{|_{least}} = AP
                       
                                  = CP - CA
                   
                                   = \sqrt 2 - 1
(  C{P^2} = O{C^2} + O{P^2} = 1 + 1 = 2 )

  And, |z - 1{|_{\max }} = BP

                             = CP + BC

                              = \sqrt 2 + 1

Further we have to find the complex number for which | z - 1 | is least i.e. Point A

 Therefore, Imaginary (y) part of A = AP\sin {45^0} = (\sqrt 2 - 1).\frac{1}{{\sqrt 2 }} = 1 - \frac{1}{{\sqrt 2 }}

 Real (x) part of A   = OP - AP\cos {45^0} = 1 - (1 - \frac{1}{{\sqrt 2 }}) = \frac{1}{{\sqrt 2 }}

 So, Complex no. associated with point A i.e. having | z - 1| least = \frac{1}{{\sqrt 2 }} + i(1 - \frac{1}{{\sqrt 2 }})



2. |z - 25i| \le 15 find the complex no. z

 (i) having least positive Argument

 (ii) having max. positive Argument

 (iii) having least modulus

 (iv) having greatest modulus

Sol. Since |z - 25i| \le 15 represents the region under the circle of radius 15 and centre at (0,25).


(i) we have to find the least positive argument of z if |z - 25i| \le 15, clearly from figure the least positive argument is of complex no. associated with point P because  OP is making the least angle from the positive direction of x-axis or real axis

  OP = \sqrt {{{25}^2} - {{15}^2}} = \sqrt {625 - 225} = 20

 or, |{z_P}| = 20

 since angle POX = angle OCP

 Therefore, \sin \theta = \frac{{OP}}{{OC}} = \frac{{20}}{{25}} = \frac{4}{5}

 and \cos \theta = \frac{3}{5}

 since, {z_p} = |{z_p}|(\cos \theta + i\sin \theta )

 Therefore, complex no. having least argument = 20 ( 3/5 + i 4/5)

                                                                                   = 4( 3 + 4i )



(ii) 

 we have to find the max positive argument of z if |z - 25i| \le 15, clearly from figure the max positive argument is of complex no. associated with point Q because  OQ is making the max angle from the positive direction of x-axis or real axis.

 similarly from above,

 OQ = 20 = |{z_q}|

 \cos \theta = \frac{3}{5}, \sin \theta = \frac{4}{5}

 {z_q} = |{z_q}|[\cos (\pi - \theta ) + i\sin (\pi - \theta )]

 or, {z_q} = |{z_q}|( - \cos \theta + i\sin \theta )

 or, {z_q} = 20\left( { - \frac{3}{5} + \frac{4}{5}i} \right)

 or, Complex no. having max. argument = 4( -3 + 4i )



(iii) 



We have to find complex no. having least modulus. Since complex no. having least modulus must have least distance from the origin, therefore clearly from figure complex no. associated with point A have least modulus.

 therefore,

 Complex no. having least modulus = 10i

(iv) Similarly complex no. having max. modulus must have max. distance from origin i.e. point B

 therefore,

 Complex no. having max modulus = 40i
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Comments

  1. Unknown10 April 2015 at 05:22

    sir,
    it was really a superb way to explain the question. i could not get how you take angle 45 to find the complex no.A?
    y= AP sin 45, from where you got this angle 45? thanx for reply

    ReplyDelete

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