Geometrical representation of a complex number is one of the fundamental laws of algebra. A complex number z = α + iβ can be denoted as a point P(α, β) in a plane called Argand plane, where α is the real part and β is an imaginary part. The value of i = − 1 \sqrt{-1} − 1
In this article, students will learn representation of Z modulus on Argand plane, polar form, section formula and many more. Solved examples and clear diagrams will help students to have a well understanding about the topic.
Table of Contents:
Representation of Z modulus on Argand Plane Argand plane consists of real axis (x – axis) and imaginary axis (y – axis).
∣ Z ∣ = ( α − 0 ) 2 + ( β − 0 ) 2 \left| Z \right|=\sqrt{{{\left( \alpha -0 \right)}^{2}}+{{\left( \beta -0 \right)}^{2}}} ∣ Z ∣ = ( α − 0 ) 2 + ( β − 0 ) 2
= α 2 + β 2 =\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}} = α 2 + β 2
= R e ( z ) 2 + I m g ( z ) 2 =\sqrt{Re(z)^2 + Img(z)^2} = R e ( z ) 2 + I m g ( z ) 2
∣ z ∣ = ∣ α + i β ∣ = α 2 + β 2 \left| z \right|=\left| \alpha +i\beta \right|=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}} ∣ z ∣ = ∣ α + i β ∣ = α 2 + β 2
Conjugate of Complex Numbers on argand plane Conjugate of a complex number is the number with an same real part and opposite sign of imaginary part but equal in magnitude.
Let Z = α + iβ be the complex number.
Mirror image of Z = α + iβ along real axis will represent conjugate of given complex number.
Z = α + i β Z=\alpha +i\beta Z = α + i β
Z ‾ = α − i β \overline{Z}=\alpha -i\beta Z = α − i β
Distance between Two Points in Complex Plane If two complex numbers Z1 and Z2 are denoted as P and Q respectively in argand plane, then distance between P and Q will be given as,
P Q = ∣ z 2 − z 1 ∣ PQ=\left| {{z}_{2}}-{{z}_{1}} \right| P Q = ∣ z 2 − z 1 ∣ = ∣ ( α 2 − α 1 ) + i ( β 2 − β 1 ) ∣ =\left| \left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)+i\left( {{\beta }_{2}}-{{\beta }_{1}} \right) \right| = ∣ ( α 2 − α 1 ) + i ( β 2 − β 1 ) ∣ = ( α 2 − α 1 ) 2 + ( β 2 − β 1 ) 2 =\sqrt{{{\left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)}^{2}}+{{\left( {{\beta }_{2}}-{{\beta }_{1}} \right)}^{2}}} = ( α 2 − α 1 ) 2 + ( β 2 − β 1 ) 2
For example, Find the distance of a point P, Z = (3 + 4i) from origin.
Solution:
|Z| = |3 + 4i|
= 3 2 + 4 2 = 5 =\sqrt{{{3}^{2}}+{{4}^{2}}}=5 = 3 2 + 4 2 = 5
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Polar form of complex number Complex numbers can be represented in both rectangular and polar coordinates. Here we will study about the polar form of any complex number.
Z = ( α + i β ) Z=\left( \alpha +i\beta \right) Z = ( α + i β )
Let ‘θ’ be the angle between ‘OP’ and real axis.
Z = α + i β , ∣ z ∣ = r Z=\alpha +i\beta ,\,\,\,\left| z \right|=r Z = α + i β , ∣ z ∣ = r OM = r cos θ = α
ON = r sin θ = β
Hence, Z = α + i β Z=\alpha +i\beta Z = α + i β
= (OM) + i (ON)
= r cos θ + i r sin θ =r\cos \theta +i\,\,r\sin \theta = r cos θ + i r sin θ = r ( cos θ + i sin θ ) =r\left( \cos \theta +i\,\,\sin \theta \right) = r ( cos θ + i sin θ ) r = α 2 + β 2 = ∣ z ∣ = ∣ α + i β ∣ r=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=\left| z \right|=\left| \alpha +i\beta \right| r = α 2 + β 2 = ∣ z ∣ = ∣ α + i β ∣ Where ‘θ’ is called argument of complex number.
θ = arg ( z ) \theta =\arg \left( z \right) θ = arg ( z ) For conjugate of complex number
arg ( z ‾ ) = − θ \arg \left( \overline{z} \right)=-\theta arg ( z ) = − θ
Algebra of Complex Numbers Let’s discuss the different algebras of complex numbers. Mainly we deal with addition, subtraction, multiplication and division of complex numbers.
Product of two complex number
Let P and Q represent complex number
Z 1 = ( α 1 + i β 1 ) {{Z}_{1}}=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right) Z 1 = ( α 1 + i β 1 )
Z 2 = ( α 2 + i β 2 ) {{Z}_{2}}=\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right) Z 2 = ( α 2 + i β 2 )
θ 1 = arg ( z 1 ) {{\theta }_{1}}=\arg \left( {{z}_{1}} \right) θ 1 = arg ( z 1 )
θ 2 = arg ( z 2 ) {{\theta }_{2}}=\arg \left( {{z}_{2}} \right) θ 2 = arg ( z 2 )
Now, on multiplying Z1 Z2 = Z
Z = ( α 1 + i β 1 ) . ( α 2 + i β 2 ) Z=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right).\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right) Z = ( α 1 + i β 1 ) . ( α 2 + i β 2 ) = r 1 ( cos θ 1 + i sin θ 1 ) . r 2 ( cos θ 2 + i sin θ 2 ) ={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right).\,{{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right) = r 1 ( cos θ 1 + i sin θ 1 ) . r 2 ( cos θ 2 + i sin θ 2 ) = r 1 r 2 [ cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ] ={{r}_{1}}{{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right] = r 1 r 2 [ cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ] Let say r 1 . r 2 = r {{r}_{1}}.\,{{r}_{2}}=r r 1 . r 2 = r
Z = r ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ) Z=r\left( \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right) Z = r ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ) Division of two complex number
Let Z 1 = α 1 + i β 1 = r 1 ( cos θ 1 + i sin θ 1 ) {{Z}_{1}}={{\alpha }_{1}}+i{{\beta }_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right) Z 1 = α 1 + i β 1 = r 1 ( cos θ 1 + i sin θ 1 )
Z 2 = α 2 + i β 2 = r 2 ( cos θ 2 + i sin θ 2 ) {{Z}_{2}}={{\alpha }_{2}}+i{{\beta }_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right) Z 2 = α 2 + i β 2 = r 2 ( cos θ 2 + i sin θ 2 ) Where, θ 1 = arg ( Z 1 ) {{\theta }_{1}}=\arg \left( {{Z}_{1}} \right) θ 1 = arg ( Z 1 )
θ 2 = arg ( Z 2 ) {{\theta }_{2}}=\arg \left( {{Z}_{2}} \right) θ 2 = arg ( Z 2 ) Now, Z represent a complex number
Z = Z 2 Z 1 = Z 2 Z 1 − 1 Z=\frac{{{Z}_{2}}}{{{Z}_{1}}}={{Z}_{2}}Z_{1}^{-1} Z = Z 1 Z 2 = Z 2 Z 1 − 1
Z = Z 2 Z 1 − 1 = Z 2 Z 1 ‾ ∣ Z ∣ 2 Z={{Z}_{2}}Z_{1}^{-1}=\frac{{{Z}_{2}}\overline{{{Z}_{1}}}}{{{\left| Z \right|}^{2}}} Z = Z 2 Z 1 − 1 = ∣ Z ∣ 2 Z 2 Z 1
= r 2 r 1 ( cos ( θ 2 − θ 1 ) + i sin ( θ 2 − θ 1 ) ) =\frac{{{r}_{2}}}{{{r}_{1}}}\left( \cos \left( {{\theta }_{2}}-{{\theta }_{1}} \right)+i\,\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right) \right) = r 1 r 2 ( cos ( θ 2 − θ 1 ) + i sin ( θ 2 − θ 1 ) )
Note:
1. On multiplication of two complex numbers their argument is added.
θ = θ 1 + θ 2 \theta ={{\theta }_{1}}+{{\theta }_{2}} θ = θ 1 + θ 2 2. On division of two complex numbers their argument is subtracted.
θ = θ 1 − θ 2 \theta ={{\theta }_{1}}-{{\theta }_{2}} θ = θ 1 − θ 2
Equation of Straight Line Passing through Two Complex Points In Algebra, we have studied that equation of straight line passing through two points (x1 , y1 ) and (x2 , y2 ) is
y − y 1 = y 2 − y 1 x 2 − x 1 ( x − x 1 ) y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) y − y 1 = x 2 − x 1 y 2 − y 1 ( x − x 1 ) In complex numbers
Equation of straight line passes through two points A(Z1 ) and B(Z2 ) can be represented as
Z − Z 1 = Z 2 − Z 1 Z 2 ‾ − Z 1 ‾ ( Z ‾ − Z 1 ‾ ) Z-{{Z}_{1}}=\frac{{{Z}_{2}}-{{Z}_{1}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}\left( \overline{Z}-\overline{{{Z}_{1}}} \right) Z − Z 1 = Z 2 − Z 1 Z 2 − Z 1 ( Z − Z 1 )
⇒ Z − Z 1 Z 2 − Z 1 = Z ‾ − Z 1 ‾ Z 2 ‾ − Z 1 ‾ \Rightarrow \frac{Z-{{Z}_{1}}}{{{Z}_{2}}-{{Z}_{1}}}=\frac{\overline{Z}-\overline{{{Z}_{1}}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}} ⇒ Z 2 − Z 1 Z − Z 1 = Z 2 − Z 1 Z − Z 1 Where Z ‾ \overline{Z} Z
Or
∣ Z Z ‾ 1 Z 1 Z 1 ‾ 1 Z 2 Z 2 ‾ 1 ∣ = 0 \left| \begin{matrix} Z & \overline{Z} & 1 \\ {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ \end{matrix} \right|=0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ Z Z 1 Z 2 Z Z 1 Z 2 1 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0 Which is the required equation of straight line.
Section Formula in Complex Numbers Let A(Z1 ), B(Z2 ) and C(Z) be the three points on a line.
If C divides line AB (internally) in ratio of m : n, so that
A C B C = m n \frac{AC}{BC}=\frac{m}{n} B C A C = n m
Then, Z = m Z 2 + n Z 1 m + n Z=\frac{m\,{{Z}_{2}}+n\,{{Z}_{1}}}{m+n} Z = m + n m Z 2 + n Z 1
Co-linearity condition
Let Z1 , Z2 and Z3 be the three points A(Z1 ), B(Z2 ) and C(Z3 ).
So, condition for collinearity will be
|AB| + |BC| = |AC|
Or,
∣ Z 1 Z 1 ‾ 1 Z 2 Z 2 ‾ 1 Z 3 Z 3 ‾ 1 ∣ = 0 \left| \begin{matrix} {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ {{Z}_{3}} & \overline{{{Z}_{3}}} & 1 \\ \end{matrix} \right|=0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 1 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0
Equation of circle in Complex Plane Equation of circle, whose centre be the point representing the complex number Z0 . And radius be ‘r’.
Equation of circle will be
∣ Z − Z 0 ∣ = r \left| Z-{{Z}_{0}} \right|=r ∣ Z − Z 0 ∣ = r
Equation of circle if Z1 and Z2 be the diametric end points of circle.
( Z − Z 1 ) ( Z ‾ − Z 2 ‾ ) + ( Z − Z 2 ) ( Z ‾ − Z 1 ‾ ) = 0 \left( Z-{{Z}_{1}} \right)\left( \overline{Z}-\overline{{{Z}_{2}}} \right)+\left( Z-{{Z}_{2}} \right)\left( \overline{Z}-\overline{{{Z}_{1}}} \right)=0 ( Z − Z 1 ) ( Z − Z 2 ) + ( Z − Z 2 ) ( Z − Z 1 ) = 0 Solved Examples
Example 1: Find the length of the line segment joining the points −1−i and 2+3i.
Solution:
Since the coordinates in the complex plane are (2, 3) and (−1,−1). Hence the required distance is 5.
Trick: We know that the distance between z1 and z2 is |z1 −z2 |. Therefore, the required length is |2+3i+1+i|=5.
Example 2: Let the complex numbers z 1 , z 2 {{z}_{1}},{{z}_{2}} z 1 , z 2 z 3 {{z}_{3}} z 3 z 0 {{z}_{0}} z 0 z 1 2 + z 2 2 + z 3 2 z_{1}^{2}+z_{2}^{2}+z_{3}^{2} z 1 2 + z 2 2 + z 3 2
Solution:
Let r be the circumradius of the equilateral triangle and ω the cube root of unity. Let ABC be the equilateral triangle with z 1 , z 2 {{z}_{1}},{{z}_{2}} z 1 , z 2 z 3 {{z}_{3}} z 3 O ′ ( z 0 ) {O}'({{z}_{0}}) O ′ ( z 0 ) O ’ A , O ’ B , O ’ C {O}’A,{O}’B,{O}’C O ’ A , O ’ B , O ’ C O A ’ , O B ’ , O C ′ O{A}’,O{B}’,O{C}' O A ’ , O B ’ , O C ′ O A ’ → = z 1 − z 0 = r e i θ O B ’ → = z 2 − z 0 = r e ( θ + 2 π 3 ) = r ω e i θ O C ’ → = z 3 − z 0 = r e i ( θ + 4 π 3 ) = r ω 2 e i θ z 1 = z 0 + r e i θ , z 2 = z 0 + r ω e i θ , z 3 = z 0 + r ω 2 e i θ z 1 2 + z 2 2 + z 3 2 = 3 z 0 2 + 2 ( 1 + ω + ω 2 ) z 0 r e i θ + ( 1 + ω 2 + ω 4 ) r 2 e i 2 θ = 3 z 0 2 , \overrightarrow{O{A}’}={{z}_{1}}-{{z}_{0}}=r{{e}^{i\theta }}\\ \overrightarrow{O{B}’}={{z}_{2}}-{{z}_{0}}=r{{e}^{\left(\theta +\frac{2\pi }{3} \right)}}=r\omega {{e}^{i\theta }} \\\overrightarrow{O{C}’}={{z}_{3}}-{{z}_{0}}=r{{e}^{i\,\left(\theta +\frac{4\pi }{3} \right)}}\\=r{{\omega }^{2}}{{e}^{i\theta }} \\\ {{z}_{1}}={{z}_{0}}+r{{e}^{i\theta }},{{z}_{2}}={{z}_{0}}+r\omega {{e}^{i\theta }},{{z}_{3}}={{z}_{0}}+r{{\omega }^{2}}{{e}^{i\theta }} \\z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=3z_{0}^{2}+2(1+\omega +{{\omega }^{2}}){{z}_{0}}r{{e}^{i\theta }}+ (1+{{\omega }^{2}}+{{\omega }^{4}}){{r}^{2}}{{e}^{i2\theta }}\\ =3z_{^{0}}^{2}, O A ’ = z 1 − z 0 = r e i θ O B ’ = z 2 − z 0 = r e ( θ + 3 2 π ) = r ω e i θ O C ’ = z 3 − z 0 = r e i ( θ + 3 4 π ) = r ω 2 e i θ z 1 = z 0 + r e i θ , z 2 = z 0 + r ω e i θ , z 3 = z 0 + r ω 2 e i θ z 1 2 + z 2 2 + z 3 2 = 3 z 0 2 + 2 ( 1 + ω + ω 2 ) z 0 r e i θ + ( 1 + ω 2 + ω 4 ) r 2 e i 2 θ = 3 z 0 2 ,
since 1 + ω + ω 2 = 0 = 1 + ω 2 + ω 4 1+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}} 1 + ω + ω 2 = 0 = 1 + ω 2 + ω 4
Example 3: If the centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1 + 2i, then find its perimeter.
Solution:
Let the vertices be z 0 , z 1 , … . . , z 5 {{z}_{0}},{{z}_{1}},…..,{{z}_{5}} z 0 , z 1 , … . . , z 5 ∣ z 0 ∣ = 5 |{{z}_{0}}|\,=\sqrt{5} ∣ z 0 ∣ = 5
⇒ A 0 A 1 = ∣ z 1 − z 0 ∣ = ∣ z 0 e i θ − z o ∣ = ∣ z 0 ∣ ∣ cos θ + i sin θ − 1 ∣ = 5 ( cos θ − 1 ) 2 + sin 2 θ = 5 2 ( 1 − cos θ ) = 5 2 sin ( θ / 2 ) A 0 A 1 = 5 . 2 sin ( π 6 ) = 5 ( b e c a u s e θ = 2 π 6 = π 3 ) \Rightarrow {{A}_{0}}{{A}_{1}}= |{{z}_{1}}-{{z}_{0}}|\,=\,|{{z}_{0}}{{e}^{i\,\theta }}-{{z}_{o}}| \\= |{{z}_{0}}||\cos \theta +i\sin \theta -1| \\=\sqrt{5}\,\sqrt{{{(\cos \theta -1)}^{2}}+{{\sin }^{2}}\theta } \\=\sqrt{5}\,\sqrt{2\,(1-\cos \theta )}\\=\sqrt{5}\,\,2\sin (\theta /2) \\{{A}_{0}}{{A}_{1}}=\sqrt{5}\,.\,2\sin \,\left(\frac{\pi }{6} \right)=\sqrt{5}\left( \text because \,\,\theta =\frac{2\pi }{6}=\frac{\pi }{3} \right) ⇒ A 0 A 1 = ∣ z 1 − z 0 ∣ = ∣ z 0 e i θ − z o ∣ = ∣ z 0 ∣ ∣ cos θ + i sin θ − 1 ∣ = 5 ( cos θ − 1 ) 2 + sin 2 θ = 5 2 ( 1 − cos θ ) = 5 2 sin ( θ / 2 ) A 0 A 1 = 5 . 2 sin ( 6 π ) = 5 ( b e c a u s e θ = 6 2 π = 3 π ) Similarly, A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5 = A 5 A 0 = 5 {{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}={{A}_{5}}{{A}_{0}}=\sqrt{5} A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5 = A 5 A 0 = 5
Hence the perimeter of, regular polygon is = A o A 1 + A 1 A 2 + A 2 A 3 + A 3 A 4 + A 4 A 5 + A 5 A 0 = 6 5 ={{A}_{o}}{{A}_{1}}+{{A}_{1}}{{A}_{2}}+{{A}_{2}}{{A}_{3}}+{{A}_{3}}{{A}_{4}}+{{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}\\=\,\,6\sqrt{5} = A o A 1 + A 1 A 2 + A 2 A 3 + A 3 A 4 + A 4 A 5 + A 5 A 0 = 6 5
