Here you will learn what is square root and how to find square root of complex number with examples.
Let’s begin –
How to Find Square Root of 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
\(\implies\) (a + ib) = \((x + iy)^2\)
\(\implies\) a + ib = \((x^2 – y^2)\) + 2ixy
On equating real and imaginary parts, we get
\(x^2 – y^2\) = a ………….(i)
and, 2xy = b …………..(ii)
Now, \((x^2 + y^2)^2\) = \((x^2 – y^2)\) + \(4x^2y^2\)
\(\implies\) \((x^2 + y^2)^2\) = \(a^2 + b^2\)
\(\implies\) \((x^2 + y^2)\) = \(\sqrt{a^2 + b^2}\) ………..(iii)
Solving the equations (i) and (iii), we get
\(x^2\) = \((1\over 2)\){\(\sqrt{a^2 + b^2} + a\)} and \(y^2\) = \((1\over 2)\){\(\sqrt{a^2 + b^2} – a\)}
\(\implies\) x = \(\pm\) \(\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} + a}}\) and y = \(\pm\) \(i\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} – a}}\)
If b is positive, then by equation (ii), x and y are of the same sign.
Hence, \(\sqrt{a + ib}\) = \(\pm\) [\(\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} + a}}\) + \(i\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} – a}}\)]
If b is negative, then by equation (ii), x and y are of the different signs.
Hence, \(\sqrt{a + ib}\) = \(\pm\) [\(\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} + a}}\) – \(i\sqrt{{(1\over 2)}{\sqrt{a^2 + b^2} – a}}\)]
Example : Find the square root of 7 – 24i.
Solution : Let \(\sqrt{7 – 24i}\) = x + iy. Then,
(7 – 24i) = \((x + iy)^2\)
\(\implies\) 7 – 24i = \((x^2 – y^2)\) + 2ixy
On equating real and imaginary parts, we get
\(x^2 – y^2\) = 7 ……….(i)
and, 2xy = -24 …………..(ii)
Now, \((x^2 + y^2)^2\) = \((x^2 – y^2)^2\) + \(4x^2y^2\)
\(\implies\) \((x^2 + y^2)^2\) = 49 + 576 = 625
\(\implies\) \(x^2 + y^2\) = 25 ………..(iii)
On Solving equation (i) and (iii), we get
\(x^2\) = 16 and \(y^2\) = 9
\(\implies\) x = \(\pm 4\) and y = \(\pm 3\)
From (ii), 2xy is negative. So, x and y are of opposite signs.
\(\therefore\) (x = 4 and y = -3) or, (x = -4 and y = 3)
Hence, \(\sqrt{7 – 24i}\) = \(\pm\) (4 – 3i)