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 \(-\sqrt{\frac{7}{2}}\left(\sqrt{\sqrt{10}-1}+i\sqrt{\sqrt{10}+1}\right)\).