Representations of Complex Numbers
The complex number z = x +iy can be uniquely represented by the point P(x, y) in the co-ordinate plane and conversely corresponding to the point P(x, y) in the plane there exists a unique complex number z = x +iy. The plane is called the complex plane and the representation of complex numbers as points in the plane is called Argand diagram.
Notice that length OP = x² +y² = |z|
Also note that every real number x = x +0i is represented by point (x, 0) lying on x-axis, and every purely imaginary number iy is represented by point (0, y) lying on y-axis. Consequently, x-axis is called the real axis and y-axis is called the imaginary axis.
- If z1 = x1 +iy1 and z2
= x2 +iy2 are two complex numbers, then the distance
between two corresponding points P1(x1, y1)
and P2(x2, y 2) is
|P1P2| = [(x1 -x2)² +(y1 -y2)²] = |z1 -z2| = |z2 -z1|.
- If P1(x1, y1) and P2(x
2, y2) are two points, then the point Q dividing [P1P2]
in the ratio m : n is given
Putting m = n = 1, the mid-point of z1 and z2 =
z = x +iy = r cos +i r sin
= r (cos +i sin )
This form of z is called trigonometric form or polar form. Thus if modulus of z is r and amp(z) = , then z = r (cos +i sin ) = r cis .
Notice that if x = 0, then = /2; if y > 0 and = - /2 if y < 0.
If x 0, then tan = r sin /r cos = y/x, so that
amp (z) = = tan-1(y/x)
The unique value of q such that - < is called principal value of amplitude or argument.
For example, let z = 1 +i. Then
r = [(1)² + (1)²] = 2 and = tan-1 1 = /4.
As another example, let r = 2 , = - /4
Then z = r (cos +i sin )
Frequently we have to convert the complex number z = x +i y to its polar
form z = r cis . Do the calculations as follows:
r = x² +y²
To find :
If x = 0 i.e. if z is purely imaginary, then
= /2 if y> 0, = -/2 if y < 0.
If y = 0 i.e. if z is purely real, then
= 0 if x > 0, = if x < 0.
Otherwise, let be such that 0 < < /2
Then = if x > 0, y > 0 (i.e. z is in first quadrant)
= - if x < 0, y > 0 (i.e. z is in second quadrant)
= + if x < 0, y < 0 (i.e. z is in third quadrant)
= - if x>0, y<0 (i.e. z is in fourth quadrant).
Conversely, given z = r cis , convert it to standard form z = x +iy by using x = r cos , y = r sin .
If z is any complex number, show that -|z|Re(z)|z|. When do the equality signs hold?
Let z = x +iy = r cis = r (cos +i
sin ) where r = |z| 0 and
We know that -1 cos 1 for all
=> -r r cos r (as r0)
=> -|z| Re(z) |z|. (because Re(z) = r cos )
Now -|z| = Re(z) <=> -[x²+y²] = x <=> y = 0 and x 0.
Also Re(z) = |z| <=> x = [x²+y²] <=> y = 0 and x 0.
Hence -|z| = Re(z) = |z| <=> y = 0 and x = 0 => z = 0.
Show that the area of the triangle on the Argand plane formed by the complex numbers z, iz and z +iz is |z|²/2
Note that the points O(0), P(z), R(z +iz) and Q(iz) form parallelogram OPRQ.
Also |OP| = |z| = |iz| = |OQ| and POQ = 90°
Thus 0, z, z +iz, iz form a square of side |z|.
Hence area of triangle with vertices z, iz and z +iz
= (1/2)|z|.|z| = (1/2)|z|²
- Represent the following complex numbers in polar form
(i) 3 +i
(iii) sin 90° +cos 90°
(iv) tan -i
(v) 1 +sin +i cos
(vi) -1 -i
- Represent the following complex numbers in standard form x +iy
(i) cis 2/3
- If z is a complex number, represent -z and -iz in complex plane.
- Using distance formula, prove that the points 1, (-1 +3i)/2 and (-1 -3i)/2 form the vertices of an equilateral triangle in complex plane.
- Using distance formula or otherwise, prove that the points -2 +3i, -1 +2i and 2 +5i are collinear.
[Hint. Let the points A, B, C represent the complex numbers -2 +3i,-1 +2i and 2 +5i respectively. Show that AB +BC = AC.]
- If z1, z2, z3, z4 are complex
numbers, show that they are vertices of a parallelogram in the Argand diagram
if and only if z1 +z3 = z2 +z4.
[Hint. What is mid-point of AC? BD?]
Answers1. (i) 2 cos/6 (ii) cis 0 (iii) cis 0
(vi) 2 cis 3/4
2. (i)(1/2) + [(3)/2] i (ii) 1 -i