Solved questions by using geometrical representation of complex number

By -
LIKE TO JOIN ON FACEBOOK (click here)
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
LIKE TO JOIN ON FACEBOOK (click here)

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

Post a Comment

Popular posts from this blog

Properties of Modulus & Argument: Complex Number

By -
Image
Properties of Modulus: Modulus of z is the length of vector representing z form origin to the point z. 1.  If z = x + iy then   2.   &  Since,  ( equality follows when y = 0) and   (equality follows when x = 0) IMP 3.  Since, and,  4.  ,  5.  6.  7.   holds when  i.e. both the vectors are in same direction.  holds when   i.e. both the vectors are in opposite direction In general, 8.  IMP 9.  Explanation: Since,  Therefore,  or,                                                  [Since   (? click here ) ] and  10.     ( Property of Parallelogram ) Properties of Argument: Argument is the angle between the vector representing z from the positive direction of x-axis The Principal Argument belongs to (-pi, pi] 1.  ,   2.   , where k = 0, 1 or -1 3.   , where k = 0, 1 or -1 4.   , z cannot be negative Real Number because for negative Real number Arg(z) is equal to pi and so   will become -pi but i
Read more

Fractional Part function

By -
Image
10. Fractional Part f(x) = {x}  Every number 'x' can be written as the sum of its integer and fractional parts. x = [x] + {x} Therefore fractional part, {x} = x - [x] For example, 4.7 = 4 + 0.7 or, 0.7 = 4.7 - 4 i.e. fractional part = Number - Integral Part x = 4.7 , [x] = [4.7] = 4 and {x} = {4.7} = 0.7 Consider, {-4.7} {-4.7} = -4.7 - [-4.7] = -4.7 - (-5)  or, {-4.7} = 0.3 if the number is integer, its fractional part is obviously zero. {6} = 6 - [6] = 6 - 6 = 0 Note- fractional part is always positive and it never becomes 1. in other words, fractional part of a number is the difference between the number and its integral part. Remember,  i.e. fractional part is always less than 1 and greater than or equal to zero. When x will be an integer fractional part will become zero i.e. integer has no fractional part. How to draw the graph, note this, , if  because the fractional part is equal to the number if number is between
Read more

Geometrical Meaning of Argument and Modulus - I: Complex Number

By -
Image
LIKE TO JOIN ON FACEBOOK ( click here ) Students are requested to understand this carefully   1. Arg(z) = pi/3 z = 0 is excluded since z = 0 is the only complex number whose argument is not defined 2. Arg(z) > pi/3 Since principal argument belongs to (-pi, pi],  therefore Argument cannot be greater than pi. therefore Arg(z) > pi/3 =>)  z = 0 is excluded 3. |Arg(z)| < pi/4 => -pi/4 < Arg(z) < pi/4 z = 0 is excluded 4. => Arg(z) = pi since, principal argument belongs to  , therefore Argument cannot be greater than pi. z = 0 is excluded 5.  -pi/3 is included z = 0 is excluded 6. Complex number   , equal only when z is purely real number or '0' IMP 7. Arg(|z|) = 0 Since |z| is a Positive Real number , therefore it lies on right side of x - axis i.e. on positive Real axis and x-axis has '0' argument because it makes 0 radian angle from x-axis For z = 0 , argument does not exist T
Read more