# Properties of Conjugate , Modulus & Argument

∎ $\displaystyle \bar{\bar{z}} = z$

∎$\displaystyle |z| = |\bar{z}|$

∎ $\displaystyle z + |z| = 2 Re(z)$

∎  $\displaystyle z – \bar{z} = 2i Im(z)$

∎ If z is purely real $\large z = \bar{z}$

∎ If z is purely imaginary $\large z = -\bar{z}$

∎ $\large z \bar{z} = |z|^2 = |\bar{z}|^2$

∎$\large \overline{z_1 + z_2} = \bar{z_1}+\bar{z_2}$

In general,
$\large \overline{z_1 + z_2 + …+ z_n} = \bar{z_1}+\bar{z_2} + …+ \bar{z_n}$

∎$\large \overline{z_1 – z_2} = \bar{z_1}-\bar{z_2}$

∎ $\large \overline{z_1 . z_2} = \bar{z_1}. \bar{z_2}$

In general

$\large \overline{z_1 . z_2 …z_n} = \bar{z_1}.\bar{z_2} …\bar{z_n}$

∎$\large \bar{z^n} = (\bar{z})^n$

∎$\large \overline{(\frac{z_1}{z_2})} = (\frac{\bar{z_1}}{\bar{z_2}})$

∎ If α = f(z), then $\large \bar{ \alpha } = \bar{f(z)} = f(\bar{z})$

where α = f(z) is a function in complex variable with real coefficients.

In other words if f(x + iy) = a + ib then f(x – iy) = a – ib.

Explanation:

Let f(z) = ao + a1z + a2z2 + a3z3 + . . . + anzn, where ao , a1, a2, . . . , an are real numbers and z is a complex number. Then

$\large f(\bar{z}) = a_0 + a_1 \bar{z} + a_2 (\bar{z})^2 + a_3 (\bar{z})^3 + …. + a_n (\bar{z})^n$

$\large = \overline{a_0 + a_1 z + a_2 z^2 + a_3 z^3 + …+ a_n z^n}$

Properties of Modulus :

∎ |z| = 0 ⇒ z = 0 + i0

∎ |z1 – z2 | denotes the distance between z1 and z2

∎ -|z| ≤ Re(z) ≤ |z| ; equality holds on right or on left side depending upon z being positive real or negative real.

∎ -|z| ≤ Im z ≤ |z| ; equality holds on right side or on left side depending upon z being purely imaginary and above the real axes or below the real axes.

∎ |z| ≤ |Re(z)| + |Im(z)| ≤ √2 |z| ; equality holds on left side when z is purely imaginary or purely real and equality holds on right side when |Re(z)| = |Im(z)|

∎ |z|2 = z z

∎ |z1z2| = |z1| |z2|

In general ,

|z1 z2 …. zn| = |z1| |z2| …. |zn|

∎ |zn| = |z|n , n ∈ I

∎ $\large |\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$

∎ |z1 + z2| ≤ |z1| + |z2| => |z1 + z2 + … + zn| ≤ |z1| + |z2| + … + |zn| ; equality holds if origin, and the points represented by z1, z2, z3, …, zn are collinear and are on the same side of the origin.

∎ |z1 – z2| ≥ ||z1| – |z2|| ; equality holds when arg(z1/z2) = 0 i.e. origin and represented by z1, z2 are collinear and are on the same side of the origin.

∎ |z1 + z2|2 = (z1 + z2) ( z1 + z2 ) = |z1|2 + |z2|2 + z1 z2 + z2 z1 = |z1|2 + |z2|2 + 2Re(z1z2 )

∎ |z1 – z2|2 = (z1 – z2) ( z1 – z2) = |z1|2 + |z2|2 – z1 z2 – z2 z1 = |z1|2 + |z2|2 – 2Re(z1 z2)

### Argument & its Properties :

∎ arg(z1z2) = θ1 + θ2 = arg(z1) + arg(z2)

∎ arg (z1/z2) = θ1 – θ2 = arg(z1) – arg(z2)

∎ arg (zn) = n arg(z), n ∈ I

Note:
∎ In the above result θ1 + θ2 or θ1 – θ2 are not necessarily the principal values of the argument of corresponding complex numbers. e.g arg(zn) = n arg(z) only shows that one of the argument of zn is equal to n arg(z) (if we consider arg(z) in the principle range)!!

(i) arg(z) = 0 , π  => z is a purely real number  => z = z

(ii) arg(z) = π/2 , -π/2  => z is a purely imaginary number => z = – z

Note that the property of argument is the same as the property of logarithm.

### Complex Numbers Represented By Vectors :

It can be easily seen that multiplication by real numbers of a complex number is subjected to the same rule as the vectors.

The addition or the subtraction of two complex numbers is also the same as the addition or the subtraction of two vectors. This fact is fundamental in theory and very useful in practice.

It should be noticed that if a number z is represented by points P and OP by a vector OP, then |z| is the length OP and arg(z) is the angle which the directed line OP makes with directed OX.
Please note that if z = x + iy and P is the point (x, y), a one-to-one correspondence exists between the number z and any of the following :

(i) the point P;

(ii) the displacement OP ;

(iii) the vector OP ( or directed length )

Any one of these three things may therefore be said to represent z, or to be represented by z.

Illustration : For any three complex numbers z1 , z2 and z3 , prove that

$\large z_1 Im(\bar{z_2} z_3) + z_2 Im(\bar{z_3} z_1 ) + z_3 Im(\bar{z_1} z_2) = 0$

Solution: As $\large Im(z) = \frac{1}{2 i}(z-\bar{z})$

=> $\large z_1 Im(\bar{z_2} z_3) + z_2 Im(\bar{z_3} z_1 ) + z_3 Im(\bar{z_1} z_2)$

$\large \frac{1}{2 i} (z_1 (\bar{z_2} z_3) – z_2 \bar{z_3}) + z_2 (\bar{z_3} z_1) – z_3 \bar{z_1}) + z_3 (\bar{z_1} z_2) – z_1 \bar{z_2}))$

$\large \frac{1}{2 i} (z_1 \bar{z_2}z_3 – z_1 z_2 \bar{z_3} + z_2 \bar{z_3} z_1 – z_2 z_3 \bar{z_1} + z_3 \bar{z_1} z_2 -z_3 z_1 \bar{z_2})$

= 0

Illustration : Consider a quadratic equation az2 + bz + c = 0 where a , b , c , are complex numbers. Find the condition that the equation has
(i) one purely imaginary root
(ii) one purely real root
(iii) two purely imaginary roots
(iv) two purely real roots

Illustration : Let z1 , z2 , z3 be three distinct complex numbers satisfying | z1 − 1| = | z2 − 1| = | z3 −1|. Let A, B, and C be the points represented in the Argand plane corresponding to z1 , z2 and z3 respectively. Prove that z1 + z2 + z3 = 3 if and only if ΔABC is an equilateral triangle.

Solution: |z1 − 1| = |z2 − 1| =|z3 −1|

⇒ The point corresponding to 1(say P) is equidistant from the points A , B and C.

⇒ P is the circumcentre of the ΔABC

Now if z1 + z2 + z3 = 3 then the point corresponding to centroid of the ΔABC is $\large \frac{z_1 + z_2 + z_3}{3} = 1$

⇒ circumcentre and centroid coincide.

⇒ Δ ABC is equilateral

Conversely if Δ ABC is equilateral, then centroid is the same as the circumcentre i.e. P.

Hence centroid $\large \frac{z_1 + z_2 + z_3}{3} = 1$

⇒ z1 + z2 + z3 = 3

Exercise :

(i) If z = z1 z2 z3 ….. zn prove that

arg z − (arg z1 + arg z2 + ….. + arg zn) = 2nπ , n ∈ I.

(ii) $\large If \; x = -5 + 2\sqrt{-4}$ , find  the value of $x^4 + 9x^3 + 35 x^2 -x + 4$

(iii) Show that $\large | (2\bar{z} +5) (\sqrt{2}-i) | = \sqrt{3} |2z + 5 |$  ;  where z is a complex number.

(iv) If z1 and z2 are two complex numbers such that |z1| = |z2| + |z1 − z1|.

Show that arg z1 − arg z2 = 2nπ , n ∈ I