# 11th Class Mathematics Complex Numbers and Quadratic Equations Notes - Mathematics Olympiads - Complex Number

Notes - Mathematics Olympiads - Complex Number

Category : 11th Class

Complex Number

Complex Numbers: "Complex number is the combination of real and imaginary number".

Definition: A number of the form$x+iy,$, where $x,y\in R$ and $i=\sqrt{-1}$ is called a complex number and (i) is called iota.

A complex number is usually denoted by z and the set of complex number is denoted by C.

$\Rightarrow C=\{x+iy:x\in R,\,Y\in R,\,i=\sqrt{-1}\}$

For example: $5+3i,$ $-1+i,$ $0+4i,$ $4+0i$ etc. are complex numbers.

Note: Integral powers of iota (i)

since $i=\sqrt{-1}$ hence we have ${{i}^{2}}=-1,$ and ${{i}^{4}}=1.$

Conjugate of a complex number: If a complex number $z=a+i\,b,$ $(a,b)\in R,$ then its conjugate is

defined as $\overline{z}=a-ib$

Hence, we have

$\operatorname{Re}(z)=\frac{z+\overline{z}}{2}$ and $\operatorname{Im}(z)=\frac{z-\overline{z}}{2i}$

$\Rightarrow$ Geometrically, the conjugate of z is the reflection or point image of z in the real axis.

e.g.

(i)         $z=3-4i$

$z=3-(-4)=3+4i$

(ii)        $z=2+5i$

$\overline{z}=2-5i$

(iii)       $\overline{z}=5i$

$\overline{z}=-5i$

• Operation (i.e. Addition and Multiplication) of Complex Number

e.g.       ${{z}_{1}}=a+ib=(a,b)$

${{z}_{2}}=c+id=(c,d)$

Then ${{z}_{1}}+{{z}_{2}}=(a+c,b+d)=(a+c)+i(b+d)$

${{z}_{1}}.{{z}_{2}}=(a,b)(c,d)=(ac-bd,bc+ad)$

${{z}_{1}}-{{z}_{2}}=(a-c,b-d)\Rightarrow {{z}_{1}}-{{z}_{2}}=(a-c,b-a)$

• Some Properties of Conjugate Number

(a)        $(\overline{z})=z$

(b)        $z+\overline{z}$ if & only if z is purely real

(c)        $z+-\overline{z}$ iff z  is purely imaginary

(d)        $z+\overline{z}=2\operatorname{Re}(z)=2p$ Real part of z

(e)        $z-\overline{z}=2i\,\,lm(z)=2i$ Imaginary part of z

(f)         $\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}\pm {{\overline{z}}_{2}}$

(g)        $\overline{{{z}_{1}}.{{z}_{2}}}={{\overline{z}}_{1}}.{{\overline{z}}_{2}}$

(h)        $\left( \frac{\overline{{{z}_{1}}}}{{{z}_{2}}} \right)=\overline{\frac{{{z}_{1}}}{{{z}_{2}}}},{{z}_{2}}\ne 0.$

(i)         $\text{f}i\,\,z=\text{f}({{z}_{1}})$ then $\overline{z}=\text{f}(\overline{{{z}_{1}}})$

(j)         $(\overline{{{z}^{4}}})={{(\overline{z})}^{4}}$

(k)${{z}_{1}}{{\overline{z}}_{2}}+{{\overline{z}}_{1}}{{z}_{2}}=2\operatorname{Re}({{z}_{1}}.{{\overline{z}}_{2}})=2.Re({{z}_{1}}.{{\overline{z}}_{2}})$

• Square Roots of a Complex Number

Let z = x + iy. Let the square root of a complex number

$z=x+iy$ is $u+iv$

i.e. $\sqrt{x+iy}=u+iv$                         ......... (1)

squaring both sides we have

$x+iy={{(u+iv)}^{2}}={{u}^{2}}+{{(iv)}^{2}}+2.u.(iv)$

$={{u}^{2}}-{{v}^{2}}+2iuv$

Equating real & imaginary part, we have

$x={{u}^{2}}-{{v}^{2}}$                                        ........ (2)

$y=2uv$                                                          ........ (3)

Now, ${{u}^{2}}+{{v}^{2}}=\sqrt{{{({{u}^{2}}+{{v}^{2}})}^{2}}+4{{u}^{2}}{{v}^{2}}}$

$=\sqrt{{{x}^{2}}+{{y}^{2}}}$                                           ........ (4)

Solving (2) & (4), we have

$2{{u}^{2}}=\sqrt{{{x}^{2}}+{{y}^{2}}+x}$

${{u}^{2}}=\sqrt{\frac{{{x}^{2}}+{{y}^{2}}}{2}}+\frac{x}{2}$

$\therefore \,\,\,u=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$

Similarly, $v=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$

From (3), we can determine the sign of xy as, if xy > 0 then x & y will be the same sign.

$\sqrt{x+iy}=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+x}{2}},+i\sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$

• If $\mathbf{xy<0}$ then square root of $\mathbf{z}\sqrt{\mathbf{x+iy}}$

$=\pm \left( \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+x}{2}}-i\sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}-x}}{2}} \right)$

In short form it can be written.

i.e.        $xy>0$

Then square root of $z=x+iy$ i.e. $\sqrt{x+iy}=\pm \left( \sqrt{\frac{\left| z \right|+x}{2}}+i\sqrt{\frac{\left| z \right|-x}{2}} \right)$

e.g. Find the square root of 7+24i

Giveen $z=\sqrt{x+iy}=\sqrt{7+24i}$

$\left| \,z\, \right|=\sqrt{{{7}^{2}}+{{(24)}^{2}}}=\sqrt{625}=25$

Square root of $z=7+24i$

$=\pm \sqrt{\frac{25+7}{2}}=i\sqrt{\frac{25-7}{2}}=\pm \left( \sqrt{\frac{32}{2}}+\sqrt{\frac{18}{2}} \right)$

$=\pm (4+i3)=(4+3)$ or$-(4+3i)$

• Polar form of a complex number

Let us consider 0 as origin & OX as x-axis & OY as Y-axis.

Let $z=x+iy$ is a complex no. It is represented as $P(x,y).$

We draw $PM\bot OX$ joining P to O (origin)

Then line OP makes positive angle q with X-axis. Then by right angled triangle PMO

$\cos \theta =\frac{x}{r}=\frac{b}{h}$

$\therefore Op=r(say)$

$x=r\,\,\cos \theta$                               (i)

Similarly,

$y=r\sin \theta$                                    (ii)

$\Rightarrow$squaring & adding (i) & (ii), we have

${{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )={{x}^{2}}+{{y}^{2}}$

$\Rightarrow {{r}^{2}}={{x}^{2}}+{{y}^{2}}$

$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ &

Dividing (ii) by (i), we have

$\frac{r\sin \theta}{r\cos \theta} =\frac{y}{x}\,\,\,\,\Rightarrow \,\,\,\,\,\,\tan \theta =\left| \frac{y}{x} \right|$

$\theta ={{\tan }^{-1}}\left| \frac{y}{x} \right|$

Here, z ran be written as

$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +\sin \theta )$

which is said to be polar form of z.

$z\equiv P(r,\theta )$

Where

It is said to be modulus of complex number &

$\theta ={{\tan }^{-1}}\left| \frac{y}{x} \right|$ is said to argument or amplitude of z.

In exponential form z can be written as $r{{e}^{i\,\,\theta }}$

$\because$complex number

$z=x+iy=r(\cos \theta +\sin \theta )=r.{{e}^{i\,\,\theta }}$ (By Euler Method)

e principal value of the argument of z. then $-\pi <\theta <\pi .$

Properties of argument of complex number:

1. $\arg ({{z}_{1}}.{{z}_{2}})=\arg ({{z}_{1}})+arg({{z}^{2}})$

1. $\arg \left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\arg ({{z}_{1}})-\arg ({{z}_{2}})$

1. $\arg \left( \frac{z}{z} \right)=2.\arg z$

1. $\arg ({{z}^{n}})=n.\arg z$

1. $\arg \overline{z}=-\arg z$

1. $\arg \,\,of\,\,are\left( \frac{{{z}_{2}}}{{{z}_{1}}} \right)=\theta$ then $\arg \left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=2K\pi -\theta .$ where $K\in l.$

• Cube Root of Unity

Let $z={{(1)}^{\frac{1}{3}}}$

$\Rightarrow {{z}^{3}}=1$

$\Rightarrow {{z}^{3}}-1=0$

$\Rightarrow (z-1)({{z}^{2}}+z+1)=0$

If $z-1=0\Rightarrow z=1$

If ${{z}^{2}}+z+1=0$

Then $z=-b\pm \frac{\sqrt{{{b}^{2}}-4ac}}{2a}$

$=\frac{-1\pm \sqrt{{{1}^{2}}-4.1.1}}{2.1}=\frac{-1\pm \sqrt{-3}}{2}$

$z=\frac{-1+\sqrt{3i}}{2},\frac{-1-\sqrt{3i}}{2}$                                                $\left[ Q\sqrt{-1}=i \right]$

Putting $w=\frac{-1+\sqrt{3i}}{2}$ and $w=\frac{-1-\sqrt{3i}}{2}$

Hence, the cube root of the unity be 1,,

• Properties of cube roots of unity

(a)        Sum of cube roots of unity is zero                         i.e.        $1+w+w=0$

(b)        Product of imaginary roots is                                i.e.        $w.{{w}^{2}}=1\Rightarrow {{w}^{2}}=1$

(c)        ${{w}^{3n}}=1$                                                                    $n\in l$

(d)        ${{w}^{3n+1}}=w$ & ${{w}^{3n+2}}={{w}^{2}}$

(e)        $\overline{\omega }={{w}^{2}},$ & ${{(\overline{\omega })}^{2}}=w$

$(\overline{\omega })w={{w}^{2}},w={{e}^{\frac{2\pi i}{3}}}$

Similarly, ${{w}^{2}}={{e}^{\frac{2\pi i}{3}}}$

• Modulus of a complex number

The modulus of a complex no. $z=x+iy$ is written as mod(z) or $\left| \,z\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$

It is also said to be the absolute value of z.

e.g.       $z=3+4i$

$\Rightarrow \left| \,z\, \right|=\sqrt{{{(3)}^{2}}+{{(4)}^{2}}}=\sqrt{9+16}=\sqrt{25}=5$

• Properties of Modulus of complex number

Properties of modulus

(a)        $\left| \,z\, \right|=\left| \,\overline{z}\, \right|$

(b)        $\left| \,z\overline{z\,} \right|={{\left| \,z\, \right|}^{2}}$

(c)        $\left| \,{{z}^{4}}\, \right|={{\left| \,z\, \right|}^{4}}$

(d)        $\left| \,{{z}_{1}}{{z}_{2}}\, \right|=\left| \,z{{\,}_{1}} \right|\,\,\left| \,{{z}_{2}}\, \right|$

(e)        $\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|,\,{{z}_{2}}\ne 0$

(f)         $\left| \,{{z}_{1}}+{{z}_{2}} \right|\le \left| \,{{z}_{1}}\, \right|+\left| \,{{z}_{2}}\, \right|$

(g)        $\left| \,{{z}_{1}}-{{z}_{2}} \right|\ge \left| \,{{z}_{1}}\, \right|-\left| \,{{z}_{2}}\, \right|$

(h)        ${{\left| \,{{z}_{1}}+{{z}_{2}} \right|}^{2}}+{{\left| \,{{z}_{1}}\,-{{z}_{2}}\, \right|}^{2}}=1\left( {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}} \right)$

(i)         ${{\left| \,{{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2.\operatorname{Re}({{z}_{1}}.{{\overline{z}}_{2}})$

(j)         ${{\left| \,{{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2.\operatorname{Re}({{z}_{1}}.{{\overline{z}}_{2}})$

The cube roots of unity lies on the unity circle & divided the circumference of the circle into three equal parts.

• Root of a Complex Number

If $z=x+iy=r(\cos \theta +i\,\,\sin \theta )$ & n is a positive integer.

$\frac{1}{{{z}^{4}}}{{=}^{\frac{1}{{{r}^{n}}}}}{{\left\{ \cos (2\text{K}\pi +\theta )+isin(2\text{K}\pi +\theta ) \right\}}^{\frac{1}{4}}}{{=}^{\frac{1}{{{r}^{n}}}}}\left( \cos \frac{2\text{K}\pi \text{+}\theta }{n}+i\sin \frac{2\text{K}\pi \text{+}\theta }{n} \right)$

where K = 0, 1, 2, ……(n - 1) e.g. find nth root of unity

$\because \,\,\frac{1}{{{1}^{n}}}={{(\cos 0{}^\circ +i\sin 0{}^\circ )}^{\frac{1}{n}}}$

$=\left\{ \begin{matrix} \cos (2\text{K}\pi \text{+0)} \\ +i\,\,\sin (2\text{K}\pi +0) \\ \end{matrix} \right\}=\cos \frac{2\text{K}\pi }{n}+i\,\sin \frac{2\text{K}\pi }{n}$

$=ei\frac{2\text{K}\pi }{n}$ where, K = 0, 1, 2, 3, 4, ....n.

Putting K = 0, 1, 2, 3, 4....$(n-1).$One by one we obtain

$\frac{1}{{{1}^{n}}}=1,\,e\frac{i2\pi }{n}\,\,,e\frac{4\pi i}{n}\,\,...e\frac{i2(n-1)\pi }{n}=1,\,a,\,{{a}^{2}},{{a}^{3}}.....{{a}^{n-1}}$ where $a={{e}^{\frac{i2\pi }{n}}}$

• Some properties of nth roots

(a)        $1+a+{{a}^{2}}+....{{a}^{n-1}}=0$

(b)        $1.\,\,a.\,\,{{a}^{2}}.....{{a}^{n-1}}={{(-1)}^{n-1}}$

• Logarithm of Complex Number

$\log (a+ib)=\frac{1}{2}\log ({{\alpha }^{2}}+{{\beta }^{2}})+i{{\tan }^{-1}}\left( \frac{\beta }{\alpha } \right)$

$\log (ib)=\log b+\frac{i\pi }{2}$

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