{"id":6522,"date":"2021-10-17T22:44:54","date_gmt":"2021-10-17T17:14:54","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6522"},"modified":"2021-11-21T18:09:25","modified_gmt":"2021-11-21T12:39:25","slug":"how-to-find-square-root-of-complex-number","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/how-to-find-square-root-of-complex-number\/","title":{"rendered":"How to Find Square Root of Complex Number"},"content":{"rendered":"
Here you will learn what is square root and how to find square root of complex number with examples.<\/p>\n
Let’s begin –<\/p>\n
Let a + ib be a complex number such that \\(\\sqrt{a + ib}\\) = x + iy, where x and y are real numbers.<\/p>\n
Then,<\/p>\n
\n\\(\\sqrt{a + ib}\\) = x + iy<\/p>\n
\\(\\implies\\) (a + ib) = \\((x + iy)^2\\)<\/p>\n
\\(\\implies\\) a + ib = \\((x^2 – y^2)\\) + 2ixy<\/p>\n<\/blockquote>\n
On equating real and imaginary parts, we get<\/p>\n
\n\\(x^2 – y^2\\) = a ………….(i)<\/p>\n
and, 2xy = b …………..(ii)<\/p>\n
Now, \\((x^2 + y^2)^2\\) = \\((x^2 – y^2)\\) + \\(4x^2y^2\\)<\/p>\n
\\(\\implies\\) \\((x^2 + y^2)^2\\) = \\(a^2 + b^2\\)<\/p>\n
\\(\\implies\\) \\((x^2 + y^2)\\) = \\(\\sqrt{a^2 + b^2}\\) ………..(iii)<\/p>\n<\/blockquote>\n
Solving the equations (i) and (iii), we get<\/p>\n
\n\\(x^2\\) = \\((1\\over 2)\\){\\(\\sqrt{a^2 + b^2} + a\\)} and \\(y^2\\) = \\((1\\over 2)\\){\\(\\sqrt{a^2 + b^2} – a\\)}<\/p>\n
\\(\\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}}\\)<\/p>\n<\/blockquote>\n
If b is positive, then by equation (ii), x and y are of the same sign.<\/p>\n
\nHence, \\(\\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}}\\)]<\/p>\n<\/blockquote>\n
If b is negative, then by equation (ii), x and y are of the different signs.<\/p>\n
\nHence, \\(\\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}}\\)]<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Find the square root of 7 – 24i.<\/p>\n
Solution<\/span><\/strong> : Let \\(\\sqrt{7 – 24i}\\) = x + iy. Then,<\/p>\n
(7 – 24i) = \\((x + iy)^2\\)<\/p>\n
\\(\\implies\\) 7 – 24i = \\((x^2 – y^2)\\) + 2ixy<\/p>\n
On equating real and imaginary parts, we get<\/p>\n
\\(x^2 – y^2\\) = 7 ……….(i)<\/p>\n
and, 2xy = -24 …………..(ii)<\/p>\n
Now, \\((x^2 + y^2)^2\\) = \\((x^2 – y^2)^2\\) + \\(4x^2y^2\\)<\/p>\n
\\(\\implies\\) \\((x^2 + y^2)^2\\) = 49 + 576 = 625<\/p>\n
\\(\\implies\\) \\(x^2 + y^2\\) = 25 ………..(iii)<\/p>\n
On Solving equation (i) and (iii), we get<\/p>\n
\\(x^2\\) = 16 and \\(y^2\\) = 9 <\/p>\n
\\(\\implies\\) x = \\(\\pm 4\\) and y = \\(\\pm 3\\)<\/p>\n
From (ii), 2xy is negative. So, x and y are of opposite signs.<\/p>\n
\\(\\therefore\\) (x = 4 and y = -3) or, (x = -4 and y = 3)<\/p>\n
Hence, \\(\\sqrt{7 – 24i}\\) = \\(\\pm\\) (4 – 3i)<\/p>\n\n\n