Euler’s Form of the
complex number
The following identity is known as Euler’s formula
eiθ = cosθ + i sinθ
Euler formula gives
the polar form z = r eiθ
Note
When performing multiplication or finding powers or roots of
complex numbers, Euler form can also be used.
Example 2.22
Find the modulus and principal argument of the following complex
numbers.
(i) √3 + i
(ii) -√3 + i
(iii) - √3 - i
(iv) √3 - i
Solution
(i) √3 + i
Modulus =
Since the complex number has the principal value √3 + i
lies in the first quadrant, has the principal value
θ = α = π/6.
Therefore, the modulus and principal argument of √3 + i are
2 and π/6 respectively.
(ii) -√3 + i
Modulus = 2 and
Since the complex number -√3 + i lies in the second
quadrant has the principal value
Therefore the modulus and principal argument of (ii) -√3 + i
are 2 and 5Ï€/6 respectively.
(iii) - √3 - i
r = 2 and α = π /6 .
Since the complex number - √3 - i lies in the third
quadrant, has the principal value,
Therefore, the modulus and principal argument of - √3 - i
are 2 and -5Ï€/6 respectively.
(iv) √3 - i
r = 2 and α = π/6
Since the complex number lies in the fourth quadrant, has the
principal value,
θ = -α = -π/6
Therefore, the modulus and principal argument of
√3 - i are 2 and - π/6 .
In all the four cases, modulus are equal, but the arguments are
depending on the quadrant in which the complex number lies.
Example 2.23
Represent the complex number (i) -1- i (ii) 1+ i√3
in polar form.
Solution
(i) Let -1- i = r(cosθ + isinθ )
We have
Since the complex number -1- i lies in the third quadrant, it has
the principal value,
Note
Depending upon the various values of k , we get various
alternative polar forms.
(ii) 1+ i√3
Therefore, the polar form of 1+ i√3 can be written as
Find the principal argument Arg z , when z = -2 / [ 1+i√3 ].
This implies that one of the values of arg z is 2Ï€/3 .
Since 2π/3 lies between -π and π, the principal argument Arg z is
2 π/3.
If z = r (cosθ + i sinθ ), then z-1 =1/r (cosθ - i sinθ ) .
If z1 = r1 (cosθ1 + i sinθ1 ) and z2 = r2 (cosθ2 + i sinθ2 ) ,
then z1 z2 = r1r2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) .
z = r (cosθ1 + i sinθ1) and
z = r (cosθ2 + i sinθ2)
⇒ z1 z2 = r1 (cosθ1 + i sinθ1 ) r2 (cosθ2 + i sinθ2 )
= r1r2 ((cosθ1 cosθ2 - sinθ1 sinθ2 ) + i (sinθ1 cosθ2 + sinθ2 cosθ1 ))
z1 z2 = r1r2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) .
arg ( z1 z2 ) = θ1 + θ2 = arg ( z1 ) + arg ( z2 ) .
Proof
Using the polar form of z1 and z2, we have
Example 2.25
Find the product in rectangular from.
Solution
Example 2.26
Find the quotient in rectangular form.
Solution
Which is in rectangular form.
Example 2.27
If z = x + iy and , show that x2 + y2 =1.
Solution
⇒ x2 + y2 =1
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