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.




(i)         \[z=3-4i\]



(ii)        \[z=2+5i\]



(iii)       \[\overline{z}=5i\]



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


e.g.       \[{{z}_{1}}=a+ib=(a,b)\]


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


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


  • 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)         \[(\overline{{{z}^{4}}})={{(\overline{z})}^{4}}\]




  • 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




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






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


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)



\[y=r\sin \theta \]                                    (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.

\[z\equiv P(r,\theta )\]


r = radius vector

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


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


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

Notes - Mathematics Olympiads - Complex Number
  30 20

You need to login to perform this action.
You will be redirected in 3 sec spinner