mathematics-class-11-unit-05-chapter-09-complex-numbers-l-9-15-hwfqn3amc0s

Complex number lecture-2

Complex number

$\mathbb{C} = \{ a+ib : a,b \in \mathbb{R} \}$

$a+ib=c+id \Leftrightarrow a=c \ and \ b=d$
$a+ib \leftrightarrow (a,b) \in \mathbb{R}^2$
Addition: $(a+ib)+(c+id)=(a+c)+i(b+d)$
Product: $(a+ib).(c+id)=(ac-bd)+i(ad+bc)$
$(\mathbb{C},+,\cdot)$ is a field

Complex number lecture-2

Remark

$i^2=-1$
$(a+i0).(b+i0)=ab+i0=ab$
$(0+ib).(0+id)=-bd+i0=-bd$
$(a+ib)(c+id)=a.(c+id)+(ib).(c+id)$
$\quad \ =ac+iad+ibc-bd \\ \quad \ =(ac-bd)+i(ad+bc)$

Complex number lecture-2

Notations

$z=a+ib$

(1) Real parts of $z: \text{Re}(z)=a$

(2) Imaginary part of $z: \text{Im}(z)=b$

(3) If Re $(z)=0$ , then we say that $z$ is purely an imaginary number.

(4) If Im $(z)=0$ , then we say that $z$ is purely real number.

Complex number lecture-2

Conjugate of complex number

Conjugate of $z=a+ib:$

$\bar z = a-ib$

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Complex number lecture-2

Conjugate of complex number

$\bar z$ is mirror image of $z$ with respect to x-axis.
e.g: $z=2+3i \quad \bar z=2-3i$
$z=5 \quad \bar z=5$

Complex number lecture-2

Properties

(1) $\ z=\bar z \ \text{iff} \ z \in \mathbb{R} \ \text{(or)} \ \text{Im}(z)=0$

Let $z=a+ib$ . Suppose $z=\bar z \ \Leftrightarrow a+ib=a-ib$

$\Leftrightarrow 2ib=0$

$\Leftrightarrow b=0$

i.e., Im $(z)=0$

Complex number lecture-2

Properties

(2) $\ z= \bar{\bar z}$

Proof: Let $z=a+ib \Rightarrow \bar z=a-ib=a+i(-b)$

$\bar z=a+i[-(-b)]$

$=a+ib$

Complex number lecture-2

Properties

(3) $\ \overline {z_1+z_2}=\bar z_1+ \bar z_2$

Proof: $z_1=a+ib \quad z_2=c+id$

$\overline {z_1+z_2}= \overline {(a+c)+i(b+d)}=(a+c)-i(b+d)$

$=a-ib+c-id$

$=\bar z_1+\bar z_2$

Complex number lecture-2

Properties

(4) $\quad$ For every $z \in \mathbb{C}$

$z \bar z$ is a non-negative real number.

$(a+ib)(a-ib)=a^2+b^2 \geq0$

Remark

$z_1z_2 \neq 0$
Then $z_1 \neq 0 \ and \ z_2 \neq 0 \ \ (??)$

Complex number lecture-2

Properties

(5) $\quad \overline {z_1.z_2}=\bar z_1.\bar z_2$

$\overline {z_1.z_2}=\overline {(a+ib)(c+id)}=(ac-bd)-i(ad+bc)$

$=(a-ib).(c-id)$

$=\bar z_1.\bar z_2$

Complex number lecture-2

Properties

(6) $\ \bar z^{-1}=(\bar z)^{-1}, z \neq 0$

Proof: $z.z^{-1}=1$

$\overline {z.z^{-1}}=1 \ \Rightarrow \bar z.\bar z^{-1}=1$

$\Rightarrow (\bar z)^{-1}=\overline z^{-1}$

Complex number lecture-2

Properties

(7) $\ \overline{ (\frac {z_1}{z_2})}=\frac{\bar z_1}{\bar z_2}, z_2 \neq 0$

Proof: From 5 and 6,

$\overline{ (\frac {z_1}{z_2})} = \overline { ({z_1. \frac {1}{z_2}})} = \bar z_1.\bar z_2^{-1}$

$= \bar z_1.(\bar z_2)^{-1}$

$\overline{ (\frac {z_1}{z_2})}=\bar z_1.(\bar z_2)^{-1}=\frac{\bar z_1}{\bar z_2}$

Complex number lecture-2

Remark

$\frac{1}{z}:=z^{-1} \ ; \ z.z^{-1}=1$

$z.\frac{1}{z}=1$

$z^{-1}=\frac{1}{z}.\frac{\bar z}{\bar z}=\frac {\bar z}{z \bar z}=\frac{a-ib}{a^2+b^2} \quad z=a+ib$

$=\frac{a}{a^2+b^2}-\frac{ib}{a^2+b^2}$

Complex number lecture-2

Properties

(8) $\ \text{Re}(z) = \frac{z+\bar z}{2} \;\ \text{Im}(z)=\frac{z-\bar z}{2i}$

$z = \text{Re}(z)+i \ \text{Im}(z)$
$z = \text{Re}(z) - i \text{Im}(z)$

Complex number lecture-2

Problem 1

If $x+iy=\sqrt {\frac{a+ib}{c+id}},$ then show that $(x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}$

Complex number lecture-2

Solution

$z=\sqrt{b} \ \rightarrow \ z^2=b$

$(x+iy)^2=\frac{a+ib}{c+id}$

L.H.S $=(x+iy)^2 \overline {(x+iy)^2}=(x+iy)^2(x-iy)^2$

$=(x^2+y^2)^2 \quad \{\because (x+iy).(x-iy)=x^2+y^2 \}$

R.H.S $=\frac{a+ib}{c+id} \frac{a-ib}{c-id}=\frac{a^2+b^2}{c^2+d^2}$

$i.e., (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}$

Complex number lecture-2

Identities

$(z_1+z_2)^2=(z_1+z_2)(z_1+z_2)$

$=z_1.(z_1+z_2)+z_2.(z_1+z_2)$

$=z_1^2+z_1z_2+z_2z_1+z_2^2$

$=z_1^2+2z_1z_2+z_2^2$

$(z_1+z_2)^3=z_1^3+3z_1^2z_2+3z_1z_2^2+z_2^3$
$z_1^2-z_2^2=(z_1-z_2)(z_1+z_2)$

Complex number lecture-2

Exercise

Let $z_1,z_2 \in \mathbb{C}$ . Show that $z_1.\bar z_2+\bar z_1.z_2$ is a real number.

Complex number lecture-2

Problem 2

If the sum and product of three complex numbers are real then which of the following is/are possible for some $z_1,z_2,z_3 \in \mathbb{C}$ :

(a) Exactly one of the three number is non-real.

(b) Exactly two of the three numbers are non-real.

(c) All three numbers are non-real.

(d) All three numbers are pure imaginary and non-zero.

Complex number lecture-2

Solution

(a) is not possible.

$z_1 \neq \bar z_1, z_2=a, z_3=b, a \ \text{and} \ b \in \mathbb{R}$

$z_1+z_2+z_3=z_1+a+b \notin \mathbb{R}$

Complex number lecture-2

Solution

(b) $z_1+z_2+z_3 \in \mathbb{R}$

$z_1.z_2.z_3 \in \mathbb{R}$

$z_1=1+i \ z_2=1-i \ z_3=1$

Yes

Complex number lecture-2

Solution

(c) Yes $\quad z_1=1-i, \ z_2=1-i, \ z_3=2i$

$z_1z_2=1^2-1-2i$

$=-2i$

(d) Exercise

No

Complex number lecture-2

Modulus of a complex number

Let $a \in \mathbb{R}$

$| a | := {\begin{cases} a & a \geq 0 \\ - a & a < 0 \end{cases} = max {a, - a}$

There is no ‘<’ in $\mathbb{C}$ :
If there is a ‘<’ relation in $\mathbb{C}$ .
- Either $i < 0$ or $i > 0$
- $i^2 > 0$ but $-1 < 0$

Complex number lecture-2

Modulus of a complex number

Let $z=a+ib$
Modulus of ‘z’:
$|z|:=\sqrt {a^2+b^2}$

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Complex number lecture-2

Examples

$z=1+i$
- $|z|:= \sqrt {1^2+1^2}=\sqrt{2}$
$z=5$
- $|z|=\sqrt {(5)^2}$ $=5$
$z=i \quad |z|=\sqrt{1^2}=1$

Thank you