When performing multiplication or finding powers or roots of complex numbers, Euler form can also be used.

**Eulerâ€™s Form of the
complex number**

The following identity is known as Eulerâ€™s formula

*e*^{i}^{Î¸}* *= cos*Î¸** *+ *i *sin*Î¸*

Euler formula gives
the polar form *z *= *r e*^{i}^{Î¸}

**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) âˆš3 + *i*

(ii) -âˆš3 + *i*

(iii) - âˆš3 - *i*

(iv) âˆš3 - *i*

**Solution**

**(i) âˆš3 + i**

Modulus =

Since the complex number has the principal value âˆš3 + *i*
lies in the first quadrant, has the principal value

Î¸ = Î± = Ï€/6.

Therefore, the modulus and principal argument of âˆš3 + *i *are
2 and Ï€/6 respectively.

**(ii) -âˆš3 + i**

Modulus = 2 and

Since the complex number -âˆš3 + *i* lies in the second
quadrant has the principal value

Therefore the modulus and principal argument of (ii) -âˆš3 + *i*
are 2 and 5Ï€/6 respectively.

(iii) - âˆš3 - *i*

r = 2 and Î± = Ï€ /6 .

Since the complex number - âˆš3 - *i* lies in the third
quadrant, has the principal value,

Therefore, the modulus and principal argument of - âˆš3 - *i*
are 2 and -5Ï€/6 respectively.

**(iv) âˆš3 - i**

r = 2 and Î± = Ï€/6

Since the complex number lies in the fourth quadrant, has the
principal value,

Î¸ = -Î± = -Ï€/6

Therefore, the modulus and principal argument of

âˆš3 - *i* are 2 and - Ï€/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*âˆš3
in polar form.

**Solution**

(i) Let -1- *i* = *r*(cosÎ¸ + *i*sinÎ¸ )

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+ iâˆš3

Therefore, the polar form of 1+ iâˆš3 can be written as

Find the principal argument Arg z , when z = -2 / [ 1+*i*âˆš3 ].

This implies that one of the values of arg z is 2Ï€/3 .

Since 2Ï€/3 lies between -Ï€ and Ï€, the principal argument Arg z is
2 Ï€/3.

**If z = r (cosÎ¸ + i sinÎ¸ ),
then z^{-1} =1/r
(cosÎ¸ - i sinÎ¸ ) .**

**If z_{1} = r_{1} (cosÎ¸_{1} + i sinÎ¸_{1} ) and z_{2} = r_{2} (cosÎ¸_{2} + i sinÎ¸_{2} ) ,**

**then z_{1} z_{2} = r_{1}r_{2} (cos (Î¸_{1} + Î¸_{2} ) + i sin (Î¸_{1} + Î¸_{2} )) .**

*z *= *r *(cos*Î¸*_{1}* *+ *i *sin*Î¸*_{1}) and

*z *= *r *(cos*Î¸*_{2}* *+ *i *sin*Î¸*_{2})

*â‡’ **z*_{1} *z*_{2} = *r*_{1} (cos*Î¸*_{1} + *i *sin*Î¸*_{1} ) *r*_{2} (cos*Î¸*_{2} + *i *sin*Î¸*_{2} )

= *r*_{1}*r*_{2} ((cos*Î¸*_{1} cos*Î¸*_{2} - sin*Î¸*_{1} sin*Î¸*_{2} ) + *i *(sin*Î¸*_{1} cos*Î¸*_{2} + sin*Î¸*_{2} cos*Î¸*_{1} ))

*z*_{1} *z*_{2} = *r*_{1}*r*_{2} (cos (*Î¸*_{1} + *Î¸*_{2} ) + *i *sin (*Î¸*_{1} + *Î¸*_{2} )) .

arg ( *z*_{1} *z*_{2} ) = *Î¸*_{1} + *Î¸*_{2 }= arg ( *z*_{1} ) + arg ( *z*_{2} ) .

**Proof**

Using the polar form of z_{1 }and z_{2}, 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 *x*^{2} + y^{2} =1.

**Solution**

â‡’ *x*^{2} + y^{2} =1

Tags : Definition, Properties, Formulas, Solved Example Problems , 12th Mathematics : UNIT 2 : Complex Numbers

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12th Mathematics : UNIT 2 : Complex Numbers : Eulerâ€™s Form of the complex number | Definition, Properties, Formulas, Solved Example Problems

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