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# How do you find the square roots of a complex number?

Before finding the square root of a complex number we must know what complex numbers are.

A complex number is the sum of a real number and an imaginary number.

It is written generally in the form of a + ib and it is represented by alphabet z

Here, both a and b are real numbers.  The value ‘a’ is called the real part and is denoted by Re ( z ).

The imaginary part is denoted by b and is represented as  Im ( z ).

The alphabet ‘i’ is known as iota and represents the imaginary part of the complex number.

Further iota is very helpful to find the square root of negative numbers.

• $\displaystyle i=\sqrt{{-1}}~~$
• $\displaystyle {{i}^{2}}=~-1~~$
• $\displaystyle {{i}^{3}}=i.{{i}^{2}}\Rightarrow i(~-1)=-i~~$

And so on.

The modulus of the complex number is the distance of the complex number represented as a point in the plane (a, ib).

Here are some important points regarding complex numbers.

• All real numbers are complex numbers but the vice versa is not true.
• All imaginary numbers are complex numbers but vice versa needn’t need to be true.
• The conjugate of a complex number  z  = a  +  ib   is  z  =  a   –  ib
• The magnitude of a complex number z = a + ib  is  | z |  = $\displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}~~$

Now, let’s discuss how can we find the square root of a complex number?

So, given below are some steps which we have to follow to find the square root of a complex number.

Firstly , suppose   x + iy  be the square root of the complex number  a  +  ib .  Then we can say,

x + iy  = √ ( a + ib)

as we have supposed  x + iy be the square root of  a + ib

Now squaring both sides we get ,

$\displaystyle {{(x+iy)}^{2}}=a+ib$

$\displaystyle {{x}^{2}}+{{(iy)}^{2}}+2(x)(iy)=a+ib$

Putting the value of  $\displaystyle {{i}^{2}}=-1$

$\displaystyle {{x}^{2}}-{{y}^{2}}+i(2xy)=a+ib$

Now, equating real and imaginary parts, we get:

$\displaystyle {{x}^{2}}-{{y}^{2}}=a$   and    2xy =  b

Now   $\displaystyle {{({{x}^{2}}+{{y}^{2}})}^{2}}={{({{x}^{2}}-{{y}^{2}})}^{2}}+4{{x}^{2}}{{y}^{2}}$

= $\displaystyle {{a}^{2}}+{{b}^{2}}$

Now we have  $\displaystyle ({{x}^{2}}+{{y}^{2}})=\sqrt{{({{a}^{2}}+{{b}^{2}})}}$ ……………….  ( 1 )

Also ,   $\displaystyle {{x}^{2}}-{{y}^{2}}=a$   ………………………………. ( 2 )

Adding equation (  1 ) and equation ( 2 ) we get ,

$\displaystyle {{x}^{2}}=1/2(\sqrt{{({{a}^{2}}+{{b}^{2}}}})+a)=h$ ( say )

Now , subtracting equation  ( 2 )  from  equation ( 1 )  we get ,

$\displaystyle {{y}^{2}}=1/2(\sqrt{{({{a}^{2}}+{{b}^{2}}}})-a)=k$ ( say )

Thus,

x  = √ h , – √ h  and  y  =  √ k , – √ k

In this way we can find the square roots of a complex number.