Complex Numbers Properties: Absolute Values and Conjugates Solved Problems (part-1)
1.Complex Numbers Properties: Absolute Values and Conjugates Solved Problems (part-1)
In Complex Numbers Properties: Absolute Values and Conjugates Solved Problems (part-1),we will solved such problems which is based on properties of complex numbers.
“Before diving into these solutions, make sure you understand the foundational concept of “Geometrical Construction of Complex Numbers on Argand Plane” and have reviewed the essential “Stereographic Projection and Inverse Points of Complex Numbers.”. This will make following today’s step-by-step solutions much easier.”
Also Read This Article:- Complex conjugate
2.Absolute Values and Conjugates Solved Problems
\text{ Illustration : } 1(a). \text{ prove that} \arg z - \arg(-z) = \pm \pi, \text{ according} \\ \text{ as } \arg z \text{ is positive or } \text{ negative.} \\ \text{ Solution : Let } z = x + iy, \text{ then} \\ -z = -x - iy, \text{ where } x, y \text{ are} \text{ real} \\ \text{ we have} \\ \arg(z) - \arg(-z) = = \tan^{ -1}(z) - \tan^{ -1}(-z) \\ = \tan^{ -1}\left(\frac{ z - z}{ 1 + (z)(-z)}\right) \\ = \tan^{ -1}\left(\frac{ 0}{ 1 - z^2}\right) \\ = \tan^{ -1} 0 \\ = \pi \ (\text{ if } z \text{ is positive}) \\ \text{ If } z \text{ is negative then} \\ \arg(z) - \arg(-z) = -\pi \\ \text{ Hence } \arg(z) - \arg(-z) = \pm \pi.\\ \text{ Illustration : } 1(b). \text{ show that} \text{ amp}(z) + \text{ amp}(\bar{ z}) = 2n\pi, n \in R \\ \text{ Solution : Let } z = x + iy \text{ then} \bar{ z} = x - iy, \text{ where } x, y \text{ are real} \\ \text{ we have} \\ \text{ amp}(z) + \text{ amp}(\bar{ z}) = \text{ amp}(z\bar{ z}) \\ = \text{ amp}[(x + iy)(x - iy)] \\ = \text{ amp}(x^2 + y^2) \\ \text{ Here } x^2 + y^2 \text{ is a positive real} \\ \text{ number, say } a, \text{ since } a \text{ is a} \\ \text{ positive real number, therefore} \\ \text{ the representative point of } a \text{ in} \text{ the Argand plane will lie} \\ \text{ on the positive side of the} \text{ real axis so the principal} \\ \text{ value of } \text{ amp}(a) \text{ is } 0 \text{ and the} \\ \text{ general value is } 2n\pi, \text{ wheres} \\ n \text{ is any integer.}\\ \text{ Hence, } \text{ amp}(z) + \text{ amp}(\bar{ z}) = 2n\pi \\ \text{ Illustration : } 2. \text{ If } |z_1| = |z_2| \text{ and } \arg z_1 + \arg z_2 = 0, \text{ show} \\ \text{ that } z_1 \text{ and } z_2 \text{ are conjugate} \text{ numbers.} \\ \text{ Solution : Given } \arg z_1 + \arg z_2 = 0 \text{ and } |z_1| = |z_2| \\ \therefore \arg z_2 = -\arg z_1 \\ \text{ since the modulus of the} \text{ one number is equal to the} \\ \text{ modulus of the other and} \text{ amplitude of the one number} \\ \text{ is equal to the negative} \text{ amplitude of the other. Hence} \\ \text{ the two numbers are conjugate} \text{ to each other.} \\ \text{ Illustration : } 3. \text{ prove that the modulus of} \text{ the ratio of any two conjugate} \\ \text{ numbers is unity.} \\ \text{ Solution : Let } z_1 = x + iy, z_2 = x - iy \\ \left|\frac{ z_1}{ z_2}\right| = \frac{ |z_1|}{ |z_2|} \\ = \frac{ |x + iy|}{ |x - iy|} \\ = \frac{ \sqrt{ x^2 + y^2}}{ \sqrt{ x^2 + (-y)^2}} \\ \Rightarrow \left|\frac{ z_1}{ z_2}\right| = 1 \\ \text{ Illustration : } 4. \text{ If the sum} \text{ and product of two} \text{ complex numbers are} \\ \text{ both real, then show that} \text{ the two numbers must} \text{ be either real or conjugate.} \\ \text{ Solution : Let } z_1 = x_1 + iy_1 \text{ and} \\ z_2 = x_2 + iy_2 \text{ be two complex} \text{ numbers.} \\ \text{ Then } z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) \\ \text{ and } z_1 z_2 = (x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1) \\ \text{ Now if } z_1 + z_2 \text{ and } z_1 z_2 \text{ are both} \text{ real, then we must have} \\ y_1 + y_2 = 0 \text{ and } x_1 y_2 + x_2 y_1 = 0 \\ \text{ From the first of these, we have} \\ y_2 = -y_1 \text{ which when substituted} \\ \text{ in the second gives } x_1 = x_2 \\ \therefore z_2 = x_2 + iy_2 = x_1 - iy_1, \text{ which is} \\ \text{ conjugate of } z_1 = x_1 + iy_1 \\ \text{ Thus if } z_1 + z_2 \text{ and } z_1 z_2 \text{ are both} \\ \text{ real, then } z_1 \text{ and } z_2 \text{ are conjugate} \text{ of each other.} \\ \text{ Conversely if } z_2 \text{ is conjugate of } z_1, \text{ then } z_2 = \bar{ z}_1 = x_1 - iy_1 \\ \therefore z_1 + z_2 = (x_1 + iy_1) + (x_1 - iy_1) = 2x_1 \text{ which is real} \\ \text{ and } z_1 z_2 = (x_1 + iy_1)(x_1 - iy_1) = x_1^2 + y_1^2 \text{ which} \\ \text{ is also real} \\ \text{ Thus if } z_1 + z_2 \text{ and } z_1 z_2 \text{ are} \\ \text{ both real, then } z_1 \text{ and } z_2 \text{ are conjugate} \text{ of each other.} \\ \text{ Hence sum and product of} \text{ two complex numbers are real} \\ \text{ then they are conjugate each} \text{ other.} \\ \text{ Illustration : } 5. \text{ If } |z_1 + z_2 + \dots + z_n| = |z_1| + |z_2| + |z_3| + \dots + |z_n| \text{ then show} \\ \text{ that the all numbers } z_1, z_2, \dots, z_n\\ \text{ Solution : If } |z_1 + z_2| = |z_1| + |z_2| \text{ and } z_1 = x_1 + iy_1, z_2 = x_2 + iy_2 \\ |z_1 + z_2| = |x_1 + iy_1 + x_2 + iy_2| = |x_1 + x_2 + i(y_1 + y_2)| \\ \Rightarrow |z_1 + z_2| = \sqrt{ (x_1 + x_2)^2 + (y_1 + y_2)^2} \\ |z_1| = |x_1 + iy_1| = \sqrt{ x_1^2 + y_1^2} \\ |z_2| = |x_2 + iy_2| = \sqrt{ x_2^2 + y_2^2} \\ |z_1| + |z_2| = \sqrt{ x_1^2 + y_1^2} + \sqrt{ x_2^2 + y_2^2} \\ \text{ then } \text{ amp}(z_1 + z_2) = \tan^{ -1} \frac{ 0}{ \sqrt{ (x_1 + x_2)^2 + (y_1 + y_2)^2}} = \tan^{ -1}(0) = \pi \\ \text{ Similarly } \text{ amp}[|z_1| + |z_2|] = \tan^{ -1}\left(\frac{ 0}{ \sqrt{ x_1^2 + y_1^2} + \sqrt{ x_2^2 + y_2^2}}\right) \\ = \tan^{ -1} 0 = \pi \\ \text{ Ampti modulus of any complex} \text{ number has real value only.} \\ \text{ similarly} \text{ amplitude } |z_1 + z_2 + z_3| = \text{ amplitude} [|z_1| + |z_2| + |z_3|] \\ \text{ Hence all numbers } z_1, z_2, \dots, z_n \\ \text{ have the same amplitudes.} \text{ Illustration : } 6. \text{ If } |z| < 1, \text{ show that} \\ \left|\frac{ 1}{ 2} \arg\left(\frac{ 1 + z}{ 1 - z}\right)\right| < \frac{ \pi}{ 2} \\ \text{ Solution : Let } z = x + iy \\ \text{ then } |z| = |x + iy| = \sqrt{ x^2 + y^2} < 1 \\ \frac{ 1 + z}{ 1 - z} = \frac{ 1 + x + iy}{ 1 - x - iy} = \frac{ (1 + x + iy)(1 - x + iy)}{ (1 - x - iy)(1 - x + iy)} \\ = \frac{ 1 - x + iy + x - x^2 + ixy + iy - ixy - y^2}{ (1 - x)^2 + y^2} \\ = \frac{ 1 + 2iy - x^2 - y^2}{ (1 - x)^2 + y^2}\\ = \frac{ 1 - x^2 - y^2}{ (1 - x)^2 + y^2} + i \frac{ 2y}{ (1 - x)^2 + y^2} \\ \arg\left(\frac{ 1 + z}{ 1 - z}\right) = \tan^{ -1}\left[\frac{ \frac{ 2y}{ (1 - x)^2 + y^2}}{ \frac{ 1 - x^2 - y^2}{ (1 - x)^2 + y^2}}\right] \\ = \tan^{ -1}\left(\frac{ 2y}{ 1 - x^2 - y^2}\right) \\ = \tan^{ -1}\left[\frac{ 2y}{ 1 - x^2 - y^2}\right] \\ = \tan^{ -1}\left(\frac{ 2y}{ 1 - x^2 - y^2}\right) \\ = \tan^{ -1}\left(\frac{ 2y}{ 1 - x^2 - y^2}\right) \\ = \tan^{ -1}\left(\frac{ 2y}{ 1 - (x^2 + y^2)}\right) \\ < \tan^{ -1}\left(\frac{ 2y}{ 1 - 1}\right) \ [|z|^2 = 1] \\ < \tan^{ -1} \infty \\ < \frac{ \pi}{ 2} \\ \Rightarrow \frac{ 1}{ 2} \arg\left(\frac{ 1 + z}{ 1 - z}\right) < \frac{ \pi}{ 2} \\ \Rightarrow \left|\frac{ 1}{ 2} \arg\left(\frac{ 1 + z}{ 1 - z}\right)\right| < \frac{ \pi}{ 2} \\ \text{ Illustration : } 7. \text{ If } |z| = 1, z \ne 1, \\ \text{ show that } z \text{ can be expressed} \\ \text{ in the form } z = \frac{ 1 + ia}{ 1 - ia}, \text{ where} \\ \text{ a is a real number.} \\ \text{ Solution : } z = \frac{ 1 + ia}{ 1 - ia} \\ |z| = \left|\frac{ 1 + ia}{ 1 - ia}\right|\\ = \frac{ |1 + ia|}{ |1 - ia|} \\ = \frac{ \sqrt{ 1 + a^2}}{ \sqrt{ 1 + a^2}} \\ |z| = 1 \\ \text{ Hence proved } z \text{ can be} \text{ expressed in the form} \\ z = \frac{ 1 + ia}{ 1 - ia} \\ \text{ Illustration : } 8. \text{ Prove that;} \\ |z| \ge \frac{ |\text{ Re } z| + |\text{ Im } z|}{ \sqrt{ 2}}, \forall z \in \mathbb{ C} \\ \text{ Solution : } |z| \ge |\text{ Re}(z)| \ge \text{ Re}(z) \quad \text{ --- } (1) \\ \text{ and } |z| \ge |\text{ Im}(z)| \ge \text{ Im}(z) \quad \text{ --- } (2) \\ \text{ Now adding } (1) \text{ and } (2), \text{ we get} \\ 2|z| \ge |\text{ Re}(z)| + |\text{ Im}(z)| \\ \Rightarrow |z| \ge \frac{ |\text{ Re}(z)| + |\text{ Im}(z)|}{ \sqrt{ 2}} \\ \text{ Illustration : } 9. \text{ Prove that;} \\ (1) \ |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2 \\ \text{ Solution : } |z_1 + z_2|^2 = (z_1 + z_2)(\bar{ z}_1 + \bar{ z}_2) \\ = (z_1 + z_2)(\bar{ z}_1 + \bar{ z}_2) \\ |z_1 + z_2|^2 = z_1 \bar{ z}_1 + z_1 \bar{ z}_2 + z_2 \bar{ z}_1 + z_2 \bar{ z}_2 \quad \text{ --- } (1) \\ \text{ and } |z_1 - z_2|^2 = (z_1 - z_2)(\bar{ z}_1 - \bar{ z}_2) \\ = (z_1 - z_2)(\bar{ z}_1 - \bar{ z}_2) \\ = (z_1 \bar{ z}_1 - z_1 \bar{ z}_2 - z_2 \bar{ z}_1 + z_2 \bar{ z}_2) \\ |z_1 - z_2|^2 = z_1 \bar{ z}_1 - z_1 \bar{ z}_2 - z_2 \bar{ z}_1 + z_2 \bar{ z}_2 \quad \text{ --- } (2) \\ \text{ Adding } (1) \text{ and } (2), \text{ we get} \\ |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2 z_1 \bar{ z}_1 + 2 z_2 \bar{ z}_2\\ = 2(z_1 \bar{ z}_1 + z_2 \bar{ z}_2) \\ |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2\left[|z_1|^2 + |z_2|^2\right] \\ [\because z\bar{ z} = |z|^2] \\ \Rightarrow |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2 \\ \text{ Illustration : } 9(ii) \left|\frac{ a - b}{ 1 - \bar{ a}b}\right| < 1 \text{ if} \\ \text{ either } |a| < 1 \text{ or } |b| < 1. \\ \text{ Solution : } \left|\frac{ a - b}{ 1 - \bar{ a}b}\right| < 1 \\ \Rightarrow \frac{ |a - b|}{ |1 - \bar{ a}b|} < 1 \quad \left[\because \left|\frac{ z_1}{ z_2}\right| = \frac{ |z_1|}{ |z_2|}\right] \\ \Rightarrow |a - b| < |1 - \bar{ a}b| \\ \Rightarrow |a - b|^2 < |1 - \bar{ a}b|^2 \\ \Rightarrow (a - b)(\bar{ a - b}) < (1 - \bar{ a}b)(\overline{ 1 - \bar{ a}b}) \\ \Rightarrow (a - b)(\bar{ a} - \bar{ b}) < (1 - \bar{ a}b)(1 - a\bar{ b}) \\ \Rightarrow a\bar{ a} - a\bar{ b} - \bar{ a}b + b\bar{ b} < 1 - a\bar{ b} - \bar{ a}b + a\bar{ a}b\bar{ b} \\ \Rightarrow a\bar{ a} + b\bar{ b} - 1 - a\bar{ a}b\bar{ b} < 0 \\ \Rightarrow |a|^2 + |b|^2 - 1 - |a|^2 |b|^2 < 0 \\ \Rightarrow |a|^2 - 1 + |b|^2 - |a|^2 |b|^2 < 0 \\ \Rightarrow |a|^2 - 1 - |b|^2(|a|^2 - 1) < 0 \\ \Rightarrow (|a|^2 - 1)(1 - |b|^2) < 0 \\ \Rightarrow |a|^2 - 1 < 0 \Rightarrow |a| < 1 \\ \text{ or } 1 - |b|^2 < 0 \Rightarrow |b| > 1 \\ \text{ Illustration : } 10. \text{ If } z_1, z_2, z_3 \\ \text{ are the vertices of an isosceles} \\ \text{ triangle, right angled at the} \\ \text{ vertex } z_2, \text{ prove that} \\ z_1^2 + 2z_2^2 + z_3^2 = 2z_2(z_1 + z_3) \\ \text{ Solution : Let the complex numbers}\\ z_1, z_2, z_3 \text{ represent the points } A, B, C \text{ respectively in the Argand} \\ \text{ diagram. } \text{ Since } \angle ABC = 90^\circ, \text{ we have} \\ \arg\left(\frac{ z_2 - z_1}{ z_2 - z_3}\right) = \frac{ \pi}{ 2} \text{ or } -\frac{ \pi}{ 2} \\ \text{ So that } \frac{ z_2 - z_1}{ z_2 - z_3} \text{ is purely imaginary.} \\ \text{ Now if a complex number } z = x + iy \text{ is purely imaginary i.e.,} \\ x = 0 \text{ then } z + \bar{ z} = 0 \\ \therefore \frac{ z_2 - z_1}{ z_2 - z_3} + \frac{ \bar{ z}_2 - \bar{ z}_1}{ \bar{ z}_2 - \bar{ z}_3} = 0 \\ \Rightarrow \frac{ z_2 - z_1}{ z_2 - z_3} = -\left(\frac{ \bar{ z}_2 - \bar{ z}_1}{ \bar{ z}_2 - \bar{ z}_3}\right) \quad \text{ --- } (1) \\ \text{ Again } BA = BC \\ \text{ so that } |z_2 - z_1| = |z_2 - z_3| \\ \Rightarrow |z_2 - z_1|^2 = |z_2 - z_3|^2 \\ \Rightarrow (z_2 - z_1)(\bar{ z}_2 - \bar{ z}_1) = (z_2 - z_3)(\bar{ z}_2 - \bar{ z}_3) \quad \text{ --- } (2) \\ \text{ Multiplying } (1) \text{ and } (2), \text{ we get} \\ \frac{ (z_2 - z_1)^2 (\bar{ z}_2 - \bar{ z}_1)}{ z_2 - z_3} = -(\bar{ z}_2 - \bar{ z}_1)(z_2 - z_3) \\ \Rightarrow [(z_2 - z_1)^2 + (z_2 - z_3)^2](\bar{ z}_2 - \bar{ z}_1) = 0 \\ \Rightarrow (z_2 - z_1)^2 + (z_2 - z_3)^2 = 0 \ [\because \bar{ z}_2 \ne \bar{ z}_1] \\ \Rightarrow z_1^2 + 2z_2^2 + z_3^2 = 2z_2(z_1 + z_3) \\ \text{ Illustration : } 11. \text{ Find the regions} \text{ of the } z\text{ -plane for which } \\ \left|\frac{ z - a}{ z + \bar{ a}}\right| < 1, = 1, > 1 \text{ where } \text{ Re}(a) > 0 \\ \text{ Solution : We have } \left|\frac{ z - a}{ z + \bar{ a}}\right| < 1, = 1, > 1 \\ \Rightarrow |z - a|^2 \lt =\gt |z + \bar{ a}|^2 \\ \Rightarrow (z - a)(\bar{ z} - \bar{ a}) \lt =\gt (z + \bar{ a})\overline{ (z+\bar{ a} )} \\ \Rightarrow (z - a)(\bar{ z} - \bar{ a}) \lt =\gt (z + \bar{ a})(\bar{ z} + a) \\ \Rightarrow z\bar{ z} - z\bar{ a} - a\bar{ z} + a\bar{ a} \lt =\gt z\bar{ z} + z\bar{ a} + \bar{ a}\bar{ z} + a\bar{ a} \\ \Rightarrow -z\bar{ a} - a\bar{ z} + a\bar{ a} \lt =\gt z\bar{ a} + \bar{ a}\bar{ z} + a\bar{ a} \\ \Rightarrow z\bar{ a} + a\bar{ z} + 2\bar{ a}a \gt =\lt 0 \\ \Rightarrow z\bar{ a} + z\bar{ a} + \bar{ a}\bar{ z} + a\bar{ z} \gt =\lt 0 \\ \Rightarrow z(a + \bar{ a}) + \bar{ z}(a + \bar{ a}) \gt =\lt 0 \\ \Rightarrow (z + \bar{ z})(a + \bar{ a}) \gt =\lt 0 \\ \Rightarrow 2x \cdot 2\text{ Re}(a) \gt =\lt 0 \\ \Rightarrow x \gt =\lt 0 \ [\because \text{ Re}(a) \text{ is positive}] \\ \text{ Thus the required regions} \text{ are the right half of the} \\ \text{ z-plane, the imaginary axis} \text{ and the left half of the z-plane} \text{ respectively.}
3.Practice Problems of Absolute Values and Conjugates Solved Problems for Students
(1.)Show that locus of z such that |z - a| \cdot |z + a| = a^2, a > 0 is a leminiscate.
(2.)Prove that \left|\frac{z - 1}{z + 1}\right|=const. and amp=const. are orthogonal circles
By solving the above questions,one can understand the Absolute Values and Conjugates Solved Problems well because the concept is well understood when you solve the questions practically.
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4.Frequently Asked Questions Related to Absolute Values and Conjugates Solved Problems
Q:1.What does the argument of any complex number z depends on ?
Ans :.Argument of any complex number z depends on that quradrant.
Q:2.How do you find argument of any complex number?
Ans:(1.) if a > 0, b > 0 (first quadrant) then argument z = \tan^{-1}\frac{b}{a}
(2.)if a < 0, b > 0 (second quadrant) then argument z = \pi - \tan^{-1}\left(\frac{b}{a}\right)
(3.)if a < 0, b < 0 (third quadrant) then argment z = -\pi + \tan^{-1}\left|\frac{b}{a}\right|
(4.)if a > 0, b < 0 (fourth quadrant) then argument z = -\tan^{-1}\left|\frac{b}{a}\right|
Q:3.Give some specific conditions of the polar form of a complex number
Ans:Let polar form of a complex number= r(\cos\theta + i\sin\theta) \\
(1) 1 = \cos 0^\circ + i\sin 0^\circ \\
(2) -1 = \cos\pi + i\sin\pi \\
(3) i = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) \\
(4) \ -i = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) \\
-i = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)
By answering the above questions,you can know about the primary terms of Absolute Values and Conjugates Solved Problems.
5.छात्र-छात्राओं से आज सवाल (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!👇
*”\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right) \dots \left(1 - \frac{1}{n}\right) को सरल करके मान बताइए।”*
(Simplify \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right) \dots \left(1 - \frac{1}{n}\right) and give the value.)
*पिछली प्रश्नोत्तरी का उत्तर*
एक प्रतिशत का आधा=\frac{1\%}{2} \\ = \frac{1}{100 \times 2} = \frac{1}{200} = 0.005
*Previos Quiz Solution*
Half of one percent=\frac{1\%}{2} \\ = \frac{1}{100 \times 2} = \frac{1}{200} = 0.005
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*
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