Complex Numbers
9.0 Square root of a complex number
9.0 Square root of a complex number
The square roots of a complex number $z=a+ib$ are given by $$\begin{equation} \begin{aligned} \pm \left[ {\sqrt {\frac{{\left| z \right| + a}}{2}} + i\sqrt {\frac{{\left| z \right| - a}}{2}} } \right]{\text{ for }}b > 0{\text{ and}} \\ \pm \left[ {\sqrt {\frac{{\left| z \right| + a}}{2}} - i\sqrt {\frac{{\left| z \right| - a}}{2}} } \right]{\text{ for }}b < 0 \\\end{aligned} \end{equation} $$
The procedure to find the square root of a complex number can be easily understood with the help of the example given below.
Question 10. Find the square root of $z = 5 + i12$.
Solution: Let us assume $$\sqrt {5 + i12} = x + iy$$ Squaring both sides, we get $$\begin{equation} \begin{aligned} 5 + i12 = {\left( {x + iy} \right)^2} \\ 5 + i12 = {x^2} - {y^2} + 2ixy \\ \Rightarrow {x^2} - {y^2} = 5\;...(1){\text{ and }}xy = 6\;...(2) \\\end{aligned} \end{equation} $$
As we know that ${(a + b)^2} = {(a - b)^2} + 4ab$, therefore $$\begin{equation} \begin{aligned} {\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + 4{x^2}{y^2} \\ {\left( {{x^2} + {y^2}} \right)^2} = 25 + 4 \times 36\quad ({\text{from (1) and (2))}} \\ {\left( {{x^2} + {y^2}} \right)^2} = 169 \\ {x^2} + {y^2} = 13\quad ...(3) \\\end{aligned} \end{equation} $$ Adding $(1)$ and $(3)$, we get $$\begin{equation} \begin{aligned} {x^2} - {y^2} + {x^2} + {y^2} = 18 \\ {x^2} = 9 \\ x = \pm 3 \\\end{aligned} \end{equation} $$ Put it in equation $(3)$, we get $$\begin{equation} \begin{aligned} \pm 3y = 6 \\ y = \pm 2 \\\end{aligned} \end{equation} $$
So possible complex numbers are $3+i2$, $3-i2$, $-3+i2$ and $-3-i2$. But only those are taken which satisfy the equation $(2)$. Therefore, the square roots of $z = 5 + i12$ are $3+i2$ and $-3-i2$.
