Question

How to find the square root of a complex number?

Answer 1

Yo ho ho !!

∆ Let the Complex Number be :
$z = x + iy$

∆ You have two equations and two variables :
${a}^{2} - {b}^{2} = x$
$2ab = y$

∆ Once you solve for ( a , b ), notice :
$z = x + iy = {a}^{2} + 2abi + {(bi)}^{2}$

And hence,
$\sqrt{z} = (a + ib)$

There you go with your desired root ✓✓

Answer 2

Step-by-step explanation:

$\\ \sqrt{a ± bi} ^{ \neg} = \sqrt[2]{1} ^{\neg} ·( \sqrt{ \frac{a + \sqrt{ ({a}^{2} + {b}^{2} ) } }{2} } ±i \sqrt{ \frac{ - a + \sqrt{ ({a}^{2} + {b}^{2} )} }{2} } )$

Slide →→→

The symbol (√x' ) denotes the Real Valued square root of the complex number

or more specifically

Surd[x, 2]

Hence, Surd[1,2] = ²√1' = ±1