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Find the square root of -7+21i

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

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