Let \(\sqrt{-7+21i}\) = x + iy
⇒ (x + iy)2 = –7 + 21i (By squaring both sides)
⇒ x2 + i2y2 + 2ixy = –7 + 21i (\(\because\) (a + b)2 = a2 + b2 + 2ab)
⇒ x2 – y2 + 2xyi = –7 + 21i (\(\because\) i2 = –1)
⇒ x2 – y2 = –7 .........(1)
And 2xy = 21 ..........(2)
(By comparing real and imaginary part of complex numbers)
\(\because\) (x2 + y2)2 = (x2 – y2)2 + 4x2y2 (\(\because\) (a + b)2 = (a – b)2 + 4ab)
= (–7)2 + (21)2 (From equation (1) and (2))
= 49 + 441
= 490
⇒ x2 + y2 = \(\sqrt{490}\) \(=7\sqrt{10}\) .......(3)
By adding equations (1) and (3), we get
2x2 = –7 + \(7\sqrt{10}\)
⇒ x2 = \(\frac{7}{2}\)(–1 + \(\sqrt{10}\))
⇒ x \(=\pm\sqrt{\frac{7}{2}(-1+\sqrt{10})}\) (\(\because \) \(\sqrt{10}\) > 1 ⇒ x is a real number)
By putting the value of x2 in equation (3), we get
y2 = \(7\sqrt{10}-\mathrm x^2\)
\(=7\sqrt{10}-\frac{7}{2}(-1+\sqrt{10})\)
\(=\frac{14\sqrt{10}+7-7\sqrt{10}}{2}\)
\(=\frac{7(\sqrt{10}+1)}{2}\)
⇒ y \(=\pm\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)
\(\because \) xy = \(\frac{21}{2}> 0\) (From equation (2))
\(\therefore\) x & y have same sign.
\(\therefore\) If x \(=\sqrt{\frac{7}{2}(-1+\sqrt{10})}\) then y \(=\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)
Or
If x \(=-\sqrt{\frac{7}{2}(-1+\sqrt{10})}\) then y \(=-\sqrt{\frac{7}{2}(1+\sqrt{10})}\)
Hence, \(\sqrt{-7+21i}\) \(=\sqrt{\frac{7}{2}(\sqrt{10}-1}\) \(+i\sqrt{\frac{7}{2}(\sqrt{10}+1)}\)
\(=\sqrt{\frac{7}{2}}(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1})\)
or \(\sqrt{-7+21i}\) \(=-\sqrt{\frac{7}{2}(\sqrt{10}-1)}-i\sqrt{\frac{7}{2}\sqrt{10}+1}\)
\(=-\sqrt{\frac{7}{2}(\sqrt{10}-1)}+i\sqrt{\sqrt{10}+1}\)
Hence, the square root of –7 + 21i is \(\sqrt{\frac{7}{2}}\left(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1}\right)\) or \(-\sqrt{\frac{7}{2}}\left(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1}\right)\).