Square Root of A Complex Number

Square Root of A Complex Number

Statement: √a + ib = ± (x + iy) where $$x=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2} \right)},\,y=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2} \right)}$$.

Proof: √a + ib = ± (x + iy) → a + ib = (x + iy)² = x² – y² + 2ixy

a = x² – y² … (1)

b = 2xy … (2)

∴ (x² + y²)² = (x² – y²)² + 4x² y² = a² + b² → x² + y² = √a² + b² … (3)

Adding (1) and (3), 2x² = √a² + b² + a → $$x=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2} \right)}$$,

Subtracting (1) from (3), 2y² = √a² + b² – a → $$y=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2} \right)}$$,

∴ √a + ib = ± (x + iy) where $$x=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2} \right)},\,y=\sqrt{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2} \right)}$$.

Example 1: Find the square root of 3 + 4i.

Solution: √3 + 4i $$=\pm \left[ \frac{\sqrt{\sqrt{{{3}^{2}}+{{4}^{2}}}+3}}{2}+i\frac{\sqrt{\sqrt{{{3}^{2}}+{{4}^{2}}}-3}}{2} \right]=\pm \left( 2+i \right)$$.

Example 2: Find the square root of 7 + 24i.

Solution: Let √7 + 24i = ±(x + iy)

If 7 + 24i = a + ib then a = 7, b = 24

$$x=\frac{\sqrt{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}}{2}=\frac{\sqrt{\sqrt{{{7}^{2}}+{{24}^{2}}}+7}}{2}=\sqrt{\frac{32}{3}}=4$$,

$$y=\sqrt{\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}}=\frac{\sqrt{\sqrt{{{7}^{2}}+{{24}^{2}}}-7}}{2}=\sqrt{\frac{18}{2}}=3$$,

∴ √7 + 24i = ±(4 + 3i).

Example 3: Find the square root of – 8 – 6i.

Solution: Let √-8-6i = ± (x + iy)

If -8-6i = a + ib then a = -8, b = -6

$$x=\frac{\sqrt{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}}{2}=\frac{\sqrt{\sqrt{64+36}-8}}{2}=\sqrt{\frac{10-8}{2}}=1$$,

$$y=\sqrt{\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}}=\frac{\sqrt{\sqrt{64+36}+8}}{2}=\sqrt{\frac{10+8}{2}}=\sqrt{9}=3$$,

∴ √-8-6i = ± (1 + 3i).