Square root of a complex number

Let a+ib be a complex number such that \sqrt{a+ib} = x+iy where x and y are real numbers. Then,

\sqrt{a+ib}=x+iy

or (a+ib)=(x+iy)^{2}

or a+ib=(x^{2}-y^{2})+2ixy

On equating real and imaginary parts, we get

x^{2}-y^{2}=a

2xy=b

Now, (x^{2}+y^{2})^{2}=(x^{2}-y^{2})^{2}+4x^{2}y^{2}

or (x^{2}+y^{2})^{2}=a^{2}+b^{2}

or (x^{2}+y^{2})=\sqrt{a^{2}+b^{2}} since x^{2}+y^{2} \geq 0

From the above, we get

x^{2}=(1/2)(\sqrt{a^{2}+b^{2}}+a) and y^{2}=(1/2)(\sqrt{a^{2}+b^{2}}-a)

which in turn implies x= \pm \sqrt {(1/2)(\sqrt{a^{2}+b^{2}}+a)}

and y=\pm \sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)}

If b is positive, then by the relation 2xy=b, x and y are of the same sign. Hence,

\sqrt{a+ib}=\pm (\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}+a)}+i\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)})

If b is negative, then by the relation 2xy=b, x and y are of different signs. Hence,

\sqrt{a+ib}=\pm (\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}+a)}-i\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)}).

Note: When you have to actually, find the square root of a particular complex number or even a complex expression, carry out the above steps and don’t just mug up  the formula and try to substitute! There are a thousands of such derivations in math, with fancy formulae, so it is better to gain a deep understanding of the proofs rather than mug up  techniques or tips or tricks !!

Try this homework now:

  1. Find the square root of 1+2i
  2. Find all possible values of \sqrt{i}+\sqrt{-i}
  3. Solve for z: z^{2}-(3-2i)z=(5i-5)

Have fun the complex way 🙂

Nalin Pithwa

 

 

 

 

 

 

 

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