Square root of a complex number « Mathematics Hothouse

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:

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

Have fun the complex way 🙂

Nalin Pithwa

This entry was written by Nalin Pithwa, posted on February 15, 2016 at 1:27 pm, filed under Complex Numbers, IITJEE Mains, KVPY. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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