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Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions

October 1, 2026 Complex Analysis(English) No Comments
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1 1.Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions
1.1 2.Conjugates in Complex Numbers: Exercise Solutions
1.2 3.Practice Questions Conjugates in Complex Numbers: Exercise Solutions for Students
1.2.1 4.Frequentaly Asked Questions Related to Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions
1.2.2 Q:1.Between what is the principal value of the argument number of a complex number?
1.2.3 Q:2.What is the argument of the complex number zero?
1.2.4 Q:3.Is the angle of a complex number unique?
1.2.5 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions

In this anticle,”Conjugates in Complex Numbers: Exercise Solutions”,we will solve questions of above topics.We will use properties of complex numbers to solve questions.

Theory (थ्योरी भाग):

Complex Analysis Chapter 1: Prove that by Contour Integration (Part 1)

Complex Analysis Chapter 1: Geometrical Construction of Complex Numbers on Argand Plane (Part 2)

Complex Analysis Chapter 1: Stereographic Projection and Inverse Points of Complex Numbers (Part 3)

Exercise (अभ्यास प्रश्न):

Complex Analysis Chapter 1: Complex Numbers Properties: Absolute Values and Conjugates Solved Problems (Part 1)

Also Read This Article:- Complex conjugate

2.Conjugates in Complex Numbers: Exercise Solutions

Example: 12. Find the locus of z, when
Example: 12(i). \arg\left[\left(\frac{z-1}{z+1}\right)\right]=\frac{\pi}{3}
Solution: \arg\left[\left(\frac{z-1}{z+1}\right)\right]=\frac{\pi}{3}
we have
\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}
=\frac{x+iy-1}{(x+1)+iy}\times\frac{(x+1)-iy}{(x+1)-iy}
=\frac{x^2+x-ixy+ixy+iy-i^2y^2-x-1+iy}{(x+1)^2-i^2y^2}
=\frac{x^2+y^2+2iy-1}{(x+1)^2+y^2}
=\frac{x^2+y^2-1}{(x+1)^2+y^2}+\frac{2iy}{(x+1)^2+y^2}
After separating into real and imaginary parts
\arg\left(\frac{z-1}{z+1}\right)=\tan^{-1}\left[\frac{\frac{2y}{(x+1)^2+y^2}}{\frac{x^2+y^2-1}{(x+1)^2+y^2}}\right]
=\tan^{-1}\left(\frac{2y}{x^2+y^2-1}\right)
Hence \tan^{-1}\left(\frac{2y}{x^2+y^2-1}\right)=\frac{\pi}{3}
\Rightarrow\frac{2y}{x^2+y^2-1}=\tan\frac{\pi}{3}
\Rightarrow\frac{2y}{x^2+y^2-1}=\sqrt{3}
\Rightarrow x^2+y^2-1=\frac{2}{\sqrt{3}}y
\Rightarrow x^2+y^2-\frac{2}{\sqrt{3}}y-1=0
which is the required locus and is a circle.
Example: 12(ii). \left|\frac{z-i}{z+i}\right|\geq2
Solution: \left|\frac{z-i}{z+i}\right|\geq2
\Rightarrow\left|\frac{z-i}{z+i}\right|^2\geq4
\Rightarrow|z-i|^2\geq4|z+i|^2
\Rightarrow(z-i)(\overline{z-i})\geq4(z+i)(\overline{z+i})
\Rightarrow(z-i)(\bar z+i)\geq4(z+i)(\bar z-i)
\Rightarrow z\bar z+zi-\bar zi-i^2\geq4(z\bar z-zi+\bar zi-i^2)
\Rightarrow z\bar z+zi-\bar zi+1\geq4(z\bar z-zi+\bar zi+1)
\Rightarrow z\bar z+zi-\bar zi+1\geq4z\bar z-4zi+4\bar zi+4
\Rightarrow3z\bar z-5zi+5\bar zi+3\leq0
\Rightarrow3(x^2+y^2)+5i(\bar z-z)+3\leq0
\Rightarrow3(x^2+y^2)+5i(-2iy)+3\leq0
\Rightarrow3x^2+3y^2-10i^2y+3\leq0
\Rightarrow3x^2+3y^2+10y+3\leq0
which represents the interior and boundary of the circle 3x^2+3y^2+10y+3 \leq 0
Example: 12(iii). |z^2-1|<1
Solution: |z^2-1|<1
\Rightarrow|z^2-1|^2<1
\Rightarrow(z^2-1)(\overline{z^2-1})<1
\Rightarrow(z^2-1)(\bar z^2-1)<1
\Rightarrow z^2\bar z^2-z^2-\bar z^2+1<1
\Rightarrow(z\bar z)^2-(z^2+\bar z^2)<0
\Rightarrow(x^2+y^2)^2-2(x^2-y^2)<0
Putting x=r\cos\theta,\ y=r\sin\theta
\Rightarrow(r^2\cos^2\theta+r^2\sin^2\theta)^2-2(r^2\cos^2\theta-r^2\sin^2\theta)<0
\Rightarrow r^4(\cos^2\theta+\sin^2\theta)-2r^2(\cos^2\theta-\sin^2\theta)<0
\Rightarrow r^4-2r^2\cos2\theta<0
\Rightarrow r^2<2\cos2\theta
which represents the interior of the curve r^2=2\cos2\theta
Example: 13. Show that
|\alpha+\sqrt{\alpha^2-\beta^2}|+|\alpha-\sqrt{\alpha^2-\beta^2}|=|\alpha+\beta|+|\alpha-\beta|
all the numbers involved being complex.
Solution: Let z_1=\alpha+i\sqrt{\alpha^2-\beta^2},
z_2=\alpha-i\sqrt{\alpha^2-\beta^2} \cdots(1)
\frac{1}{2}[|z_1+z_2|^2+|z_1-z_2|^2]=|z_1|^2+|z_2|^2 (By parallelogram law)
\Rightarrow|z_1|^2+|z_2|^2=\frac{1}{2}\left|\alpha+\sqrt{\alpha^2-\beta^2}+\alpha-i\sqrt{\alpha^2-\beta^2}\right|^2
\qquad+\frac{1}{2}\left|\alpha+i\sqrt{\alpha^2-\beta^2}-\alpha+i\sqrt{\alpha^2-\beta^2}\right|^2
=\frac{1}{2}|2\alpha|^2+\frac{1}{2}|2i\sqrt{\alpha^2-\beta^2}|^2
=2|\alpha|^2+2|\sqrt{\alpha^2-\beta^2}|^2
\Rightarrow|z_1|^2+|z_2|^2=2|\alpha|^2+2|\alpha^2-\beta^2| \cdots(2)
[|z_1|+|z_2|]^2=|z_1|^2+|z_2|^2+2|z_1||z_2|
=2|\alpha|^2+2|\alpha^2-\beta^2|+2|\alpha^2-(\alpha^2-\beta^2)|[From (1) and (2)]
=2|\alpha|^2+2|\beta|^2+2|\alpha+\beta||\alpha-\beta|
=|\alpha+\beta|^2+|\alpha-\beta|^2+2|\alpha+\beta||\alpha-\beta|(using parallelogram law)
=[|\alpha+\beta|+|\alpha-\beta|]^2
\therefore |z_1|+|z_2|=|\alpha+\beta|+|\alpha-\beta|
\Rightarrow|\alpha+i\sqrt{\alpha^2-\beta^2}|+|\alpha-i\sqrt{\alpha^2-\beta^2}|=|\alpha+\beta|+|\alpha-\beta|
Example: 14. If a and b are complex numbers, find numbers z_1 and z_2 so that z_1,z_2 and a,b are opposite vertices of a square.
Solution: Let z be the point opposite vertices of a square.
AP \bot BP
\amp \left(\frac{z-a}{z-b}\right)=\frac{\pi}{2}
\Rightarrow\frac{z-a}{z-b}+\frac{\bar z-\bar a}{\bar z-\bar b}=0\cdots(1)
Again, PAQB is a square so that AP=BP
|z-a|=|z-b|
\Rightarrow|z-a|^2=|z-b|^2
\Rightarrow(z-a)(\bar z-\bar a)=(z-b)(\bar z-\bar b)
\Rightarrow(z-a)(\bar z-\bar a)=(z-b)(\bar z-\bar b)
\Rightarrow\frac{z-a}{z-b}=\frac{\bar z-\bar a}{\bar z-\bar b}\cdots(2)
From (1) and (2), we get
\frac{z-a}{z-b}+\frac{z-b}{z-a}=0
\Rightarrow(z-a)^2+(z-b)^2=0
\Rightarrow2z^2-2z(a+b)+(a^2+b^2)=0
\Rightarrow z=\frac{2(a+b)\pm\sqrt{4(a+b)^2-8(a^2+b^2)}}{4}
\Rightarrow z=\frac{1}{2}[(a+b)\pm i(a-b)]
z_1=\frac{1}{2}[(a+b)+i(a-b)]
The above two values of z are the values of z_1 and z_2.
Example: 15. Given a point \left( \xi ,\eta ,\zeta \right) on the Riemann sphere of unit radius, find its corresponding point on the complex plane.
Solution: Let the sphere be
\xi^2+\eta^2+\zeta^2=1\cdots(1)
and the plane of projection be
\zeta=0\cdots(2)
Give \left( \xi ,\eta ,\zeta \right) be co-ordinates of any point P on the sphere and \left( x,y,0\right) the co-ordinates of the corresponding point p* where the line NP* meets the plane of projection.
Since the points (\xi,\eta,\zeta) and \left( x,y,0 \right) are in a straight line, therefore we have
\frac{\xi-0}{x-0}=\frac{\eta-0}{y-0}=\frac{\zeta-1}{0-1}=k
\Rightarrow\frac{\xi}{x}=\frac{\eta}{y}=\frac{\zeta-1}{-1}=k\cdots(3)
where k is a real number, but the point \left( \xi ,\eta ,\zeta \right) lies on a unit sphere, whose equation is given by (1), therefore
x^2k^2+y^2k^2+(1-k)^2=1
Solving for k, we obtain
k=\frac{2}{x^2+y^2+1}
In view of (3), the complex number (x,y,0) on the plane is the point
\left( \xi ,\eta ,\zeta \right)=\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)\cdots(4)
on the sphere. Rewriting (4), we identify the complex number z=x+iy with the point
\left( \xi ,\eta ,\zeta \right)=\left(\frac{z+\bar z}{|z|^2+1},\frac{z-\bar z}{i(|z|^2+1)},\frac{|z|^2-1}{|z|^2+1}\right)\cdots(5)
on the sphere. Also from (3) we have
x=\frac{\xi}{1-\zeta},\qquad y=\frac{\eta}{1-\zeta}\cdots(5)
and
z=x+iy=\frac{\xi+i\eta}{1-\zeta}\cdots(6)
Example: 16. On the Riemann sphere, what are the images of 1,-1,i \text{ and } \frac{(1-i)}{\sqrt{2}}.
Solution: Image of 1=(1,0,0)
Image of -1=(-1,0,0)
Image of i=(0,1,0)
Image of \frac{(1-i)}{\sqrt{2}}=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)
Example: 17. Show that a circle on the sphere that does not pass through the north pole corresponds to a circle in the complex plane.
Solution: Any circle on the Riemann sphere is the intersection of the unit sphere
\xi^2+\eta^2+\zeta^2=1
with some plane:
a\xi+b\eta+c\zeta+d=0\cdots(1)
In view of (5) in equation (6), the corresponding points of the plane satisfy the equation:
(x^2+y^2)+2ax+2by+d-c=0\cdots(2)
Equation (2) represents a circle if c+d\neq0 and does not pass through north pole.
Example: 18. Show that a circle on the sphere passing through the north pole Complex plane.
Solution: Equation of unit circle
\xi^2+\eta^2+\zeta^2=1\cdots(1)
with some plane:
a\xi+b\eta+c\zeta+d=0\cdots(2)
In view of (3) of question 16, if
c+d=0
then the plane (2) passes through the north pole (0,0,1).

Example: 19. Show that we may identify, by stereographic projection, the complex plane with the sphere
\xi^2+\eta^2+\left(\zeta-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2
Solution: Consider a sphere in 3D space (u,v,w) or (x,y,z) tangent to the complex plane at the origin. Set the sphere equation such that its diameter corresponds to 1 along the vertical axis, giving the shifted circle/sphere equation:
u^2+v^2+\left(w-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2
(or in standard coordinates)
x^2+y^2+\left(u-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2
By projecting a point (x,y) from the complex plane through the north pole of this sphere onto the surface (u,v,w), algebraic substitution of the line equation mapping the north pole to the plane coordinates establishes this exact implicit quadratic locus.
Example: 20. Describe the stereographic projection of points on the unit sphere x^2+y^2+z^2=1 in \mathbb{R}^3 to the extended complex plane under z=x+iy, show that under this projection, the point z=x+iy corresponds to the point
\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)
on the sphere. What corresponds on the sphere to
(i) straight lines in the z-plane,
(ii) circles in the z-plane with the origin as centre.
Solution: The line passing through N(0,0,1) and P(x,y,z) can be written in parametric form using a parameter t:
=(0,0,1)+t\left((x,y,0)-(0,0,1)\right)
Since the point P(x,y,z) lies on the unit sphere, its coordinates must satisfy the equation of the sphere.
x^2+y^2+z^2=1
Substitute the parametric forms of x,y and z:
(tx)^2+(ty)^2+(1-t)^2=1
t^2(x^2+y^2)+1-2t+t^2=1
t^2(x^2+y^2+1)-2t=0
since t\neq0 (which represents the North pole itself), we can divide by t:
t(x^2+y^2+1)=2\Rightarrow t=\frac{2}{x^2+y^2+1}
Now substitute this value of t back into the expressions for x,y and z:
x=tx\Rightarrow x=\frac{2x}{x^2+y^2+1}
y=ty\Rightarrow y=\frac{2y}{x^2+y^2+1}
z=1-t=1-\frac{2}{x^2+y^2+1}=\frac{x^2+y^2-1}{x^2+y^2+1}
Thus, the point z=x+iy corresponds to the point on the sphere:
\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)
Geometric corresponding geometric figures on the sphere
A general circle or straight line in the z-plane can be represented by the equation
a(x^2+y^2)+Bx+Cy+D=0
Using two coordinates, we can express x^2+y^2,x and y in terms of the spherical coordinates latex[/latex]:
x=\frac{X}{1+Z},\quad y=\frac{Y}{1-Z},\quad x^2+y^2=\frac{1+Z}{1-Z}
Substituting these into the general equation.
yields the intersection of a plane with the sphere
A(1+Z)+Bx+Cy+D(1-Z)=0
\Rightarrow Bx+Cy+(A-D)Z+(A+D)=0
This shows that all circles and lines in the plane correspond to circles to the sphere.
(1) Straight line passing through the origin in the z-plane.
Equation in plane:
y=mx
\Rightarrow Cy+Bx=0\qquad(\text{where }A=0\text{ and }D=0)
Equation on sphere: substituting into the plane formula gives
Bx+Cy=0
Geometric description: This is a vertical plane passing through the z-axis. Its intersection with the sphere is a great circle passing through both the North pole (0,0,1)  and the South pole (0,0,-1) .
(2) Circles in the z-plane with the origin as center.
Equation in plane:
x^2+y^2=r^2
where B=0,\quad C=0,\quad A=1,\quad D=-r^2
Equation on sphere:
\frac{1+Z}{1-Z}=r^2
\Rightarrow 1+Z=r^2-r^2Z
\Rightarrow Z(1+r^2)=r^2-1
\Rightarrow Z=\frac{r^2-1}{r^2+1}
Geometric description:
since z is a constant value between -1 and 1, this represents a plane parallel to the xy-plane. Its intersection with the sphere is a horizontal circle (a parallel of latitude) that does not pass through the poles.
Example: 21. Write the equation of a straight line joining two points z_1 and z_2 of a complex plane.
Solution: Geometric condition for collinearity
Let z be any arbitrary point lying on the straight line passing through the fixed points z_1 and z_2.
For these three points to be collinear, the vector z-z_1 must be a real multiple of the vector z_2-z_1.
Mathematically, this means the ratio of these two complex numbers must be a purely real number.
\frac{z-z_1}{z_2-z_1}=t\qquad\text{where }t\in\mathbb{R}
Using complex conjugate properties, since a number is purely real if and only if it equals its own complex conjugate z = \bar{z}, we can equate the expression to its conjugate:
\frac{z - z_1}{z_2 - z_1} = \left(\overline{\frac{z - z_1}{z_2 - z_1}}\right)
Distributing the conjugate operator over the division and subtraction gives:
\frac{z - z_1}{z_2 - z_1} = \frac{\bar{z} - \bar{z}_1}{\bar{z}_2 - \bar{z}_1}
(3) Cross multiplication: Cross multiply the denominators to eliminate the fractions:
(z - z_1)(\bar{z}_2 - \bar{z}_1) = (\bar{z} - \bar{z}_1)(z_2 - z_1)
(4) Cross-multiplication expansion: Expand both sides carefully:
z\bar{z}_2 - z\bar{z}_1 - z_1\bar{z}_2 + z_1\bar{z}_1 = \bar{z}z_2 - \bar{z}z_1 - \bar{z}_1z_2 + \bar{z}_1z_1
Since z_1\bar{z}_1 appears identically on both sides of the equation, cancel it out:
z\bar{z}_2 - z\bar{z}_1 - z_1\bar{z}_2 = \bar{z}z_2 - \bar{z}z_1 - \bar{z}_1z_2
(5) Grouping and Rearranging Move all terms to one side of the equation to match the form given in the textbook’s answers:
z\bar{z}_2 - z\bar{z}_1 - \bar{z}z_2 + \bar{z}_1z_2 + \bar{z}z_1 - \bar{z}_1\bar{z}_1 = 0
Group the terms by factoring out z and \bar{z}:
-z(\bar{z}_1 - \bar{z}_2) + \bar{z}(z_1 - z_2) + (z_1\bar{z}_2 - z_2\bar{z}_1) = 0
To exactly match the target signs, multiply the entire equation by -1:
z(\bar{z}_1 - \bar{z}_2) - \bar{z}(z_1 - z_2) + (z_2\bar{z}_1 - z_1\bar{z}_2) = 0
\implies z(\bar{z}_1 - \bar{z}_2) - \bar{z}(z_1 - z_2) + z_1\bar{z}_2 - z_2\bar{z}_1 = 0
By solving the above examples,one can understand the Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions.

3.Practice Questions Conjugates in Complex Numbers: Exercise Solutions for Students

(1.)In an Argand plane the centre of the circle |4z-8+12i|=7 has the affix
(a) 2-3i     (b) 8-12i     (c) 2+3i       (d) 8+12i
(2.)In an Argand plane the radius of the circle |5z+15-16i|=20 is
(a) 20        (b) 2            (c) 4             (d) 10
Answers: (1) (a) (2) (c)
By solving the above questions,you can understand the Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions well because the concept is well understood when you solve it practically.

4.Frequentaly Asked Questions Related to Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions

Q:1.Between what is the principal value of the argument number of a complex number?

Ans: -\pi < \theta \le \pi 

Q:2.What is the argument of the complex number zero?

Ans:Undefined

Q:3.Is the angle of a complex number unique?

Ans:No
By answering the above questions,you can know about the primary terms of Properties of Absolute Values and Conjugates in Complex Numbers: Exercise Solutions.

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”एक व्यक्ति का परिचय कराते हुए एक महिला ने कहा,”मेरी पुत्री इस व्यक्ति के पिता की एकमात्र पुत्री है।” वह व्यक्ति उस महिला से कैसे सम्बन्धित है?”*
(Introducing a man, a woman said, “My daughter is the only daughter of this man’s father.” How is the person related to the woman?)
*पिछली प्रश्नोत्तरी का हल*
माना तीसरे प्रकार की चीनी का मूल्य x रुपये प्रति किग्रा था
या \frac{5}{10} \times x + \frac{2}{10} \times 20 + \frac{3}{10} \times 30 = 40 
या \implies \frac{x}{2} = 40 - 4 - 9 = 27 \implies x = 54 
C की धनराशि
=54 रुपये
*Previous Quiz Solution*
Let the price of the third type of sugar be Rs. x per kg
or \frac{5}{10} \times x + \frac{2}{10} \times 20 + \frac{3}{10} \times 30 = 40 
or \implies \frac{x}{2} = 40 - 4 - 9 = 27 \implies x = 54 
C’s amount
=Rs. 54
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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