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.
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
)
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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
= amp(z).
We know that -1
cos
1 for all ![]()
=> -r
r cos
r
(as r
0)
=> -|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
