October 8, 2024

How to find Square root of a complex number

How to find Square root of a complex number

A number of the form z=x+iy  where x,y \in \mathbb{R}

and i=\sqrt{-1} is called a complex number where x
is called as real part and y is called imaginary part of complex number. To find the square root of a complex number, we will assume that \sqrt{x+iy}=a+ib
. Then square  both the sides  and compare real and imaginary part to  find the value of a and b
, which will give us the square root.

Formula for finding square root of a complex number:

The square root of z=x+iy is

\sqrt{x+iy}= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}+i\sqrt{\frac{\left | z \right |-x}{2}} \right ]

      for  y>0    and

= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}-i\sqrt{\frac{\left | z \right |-x}{2}} \right ]

         for       y< 0

Proof. Let square root of x+iy \, \, is \, \, a+ib

. That is  \sqrt{x+iy}=a+ib  where a
and b in \mathbb{R}
. Now square both the sides we get

x+iy= (a+ib)^{2}=a^{2}-b^{2}+2iab

Equating real and imaginary parts we get  a^{2}-b^{2}=x.....(i)\, \,\, and\,\,\, 2ab=y......(ii)

Now (a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2}+4a^{2}b^{2}\Rightarrow a^{2}+b^{2}=\sqrt{x^{2}+y^{2}} ............(iii)

Solving (i) and (iii)  we get    2a^{2}=(\sqrt{x^{2}+y^{2}})+x=\left | z \right |+x

a^{2}=\frac{\left | z \right |+x}{2}\Rightarrow a=\pm \sqrt{\frac{\left | z \right |+x}{2}}

.

Similarly b=\pm \sqrt{\frac{\left | z \right |-x}{2}}. Since 2ab=y , it is clear that both a

and b have the same sign when y
is positive and a and b
  have different sign when y is negative.  Therefore

\sqrt{x+iy}= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}+i\sqrt{\frac{\left | z \right |-x}{2}} \right ]

      for  y>0    and

= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}-i\sqrt{\frac{\left | z \right |-x}{2}} \right ]

         for       y< 0

Example : Find the square root of (-5-12i) .

Sol.  Here \left | (-5-12i)\right |= \sqrt{(-5)^{2}+(-12)^{2}}=\sqrt{25+144}=13

Applying the formula for square root we get

\sqrt{(-5-12i)}=\pm \left [ \sqrt{\frac{\left | z \right |+x}{2}} - i \sqrt{\frac{\left | z \right |-x}{2}}\right ]     (\because b

is negative)

\, \, \, \, \, \, \, \, \, \, =\pm \left ( \sqrt{\frac{13-5}{2}}-i\sqrt{\frac{13+5}{2}} \right )

=\pm (2-3i)

Trick to find the square root of a complex number: 

To find \sqrt{x+iy}  , follow the following steps:

  1. First find the number \frac{\left | y \right |}{2} .
  2.  Now factorise the given number in such a way that difference of square of these factors is equal to the real number (x).

Ex.  Find the square root of 7+4i.

Sol. Find the number \frac{\left | y \right |}{2}

  which is Equal to 12. Now factor 12 in such a way that difference of square of these factors is equal to the real number  x=7.

12= 4×3   and 4^{2}-3^{2}=16-9=7

. Therefore \sqrt{7+4i}=\pm (4+3i).

Ex.   Find the square root of (-6+8i)

.

Sol. Here \frac{\left | y \right |}{2}=\frac{8}{2}=4   and 4=\sqrt{2}\times \sqrt{8}

such that (\sqrt{2})^{2}-(\sqrt{8})^{2}=2-8=-6

Thus    \sqrt{(-6+8i)}=\pm (\sqrt{2}+\sqrt{8}i)=\pm (\sqrt{2}+2\sqrt{2}i)

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