Square Root Of a Complex Number

Generally, the square root of a complex number \(Z = a + ib\) is given by the formulae : \[\large \sqrt{a + ib} = \begin{cases} \pm \left ( \sqrt{\dfrac{|Z| + a}{2}} + i \sqrt{\dfrac{|Z| - a}{2}} \right ) \qquad , \text{If } b > 0 \\ \pm \left ( \sqrt{\dfrac{|Z| + a}{2}} - i \sqrt{\dfrac{|Z| - a}{2}} \right ) \qquad , \text{If } b < 0 \\ \end{cases}\] where, \(|Z| = \sqrt{a^2 + b^2}\)

But, it is not easy to remember this formulae and it is also confusing. But there is a very easy trick to find the square root of a complex number. So, in this note I will explain you on how to find square root of a given complex number easily without using the above formulae. I will explain it through different examples

Example 1 : Find the value of $\large \sqrt{{\color{#D61F06}5} + {\color{#3D99F6}12}i}$

Step - 1 : Find the real part and imaginary part of the given complex number. $\large \underbrace{\color{#D61F06}5}_{\text{Real Part}} \qquad + \underbrace{{\color{#3D99F6}12}}_{\text{Imaginary Part}}i$ Step - 2 : Take half of the co-efficient of the imaginary part.
$\large \sqrt{5 + {\color{#3D99F6}12}i} \implies \dfrac{\color{#3D99F6}12}{2} = \color{#E81990}6$ Step - 3 : Express the obtained number as a product of two numbers. $\large {\color{#E81990}6} = \begin{cases} 6 \times 1 \\ 3 \times 2 \\ 2 \times 3 \\ 1 \times 6 \\ \end{cases}$ Step - 4 : Find the difference of squares of the factors in each case. Select the pair of factors whose difference in squares is equal to the real part of given complex number. $\large {\color{#E81990}6} = \begin{cases} 6^2 - 1^2 = 35 \\ \boxed{\color{#D61F06}3^2 - 2^2 = 5} \\ 2^2 - 3^2 = -5 \\ 1^2 - 6^2 = -35 \\ \end{cases}$ So, the pair of numbers we got are $(3, 2)$

Step - 5 : The first number number of the selected pair is the real part and the second number of the pair is the imaginary part of the square root of the given complex number. Don't forget to keep $\pm$ sign. You should note that the sign of imaginary parts of the complex number and its square root complex number are same. Here, the imaginary part of given complex is positive so it's square root imaginary part will also be positive. $\large \therefore \sqrt{{\color{#D61F06}5} + {\color{#3D99F6}12i}} = \pm ({\color{#20A900}3} + {\color{#20A900}2}i)$

Example 2 : Find the value of $\large \sqrt{{\color{#D61F06}5} - {\color{#3D99F6}12}i}$.

The method is same, the steps are same and the values you will get after you express half of the imaginary part as difference of squares of two factors will also be same. The only difference between the before problem and this one is that the imaginary part has negative sign. So, the imaginary part of the square root will also have negative sign. $\large \dfrac{\color{#3D99F6}12}{2} = {\color{#E81990}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies \boxed{\color{#D61F06}3^2 - 2^2 = 5} \\ 2 \times 3 \implies 2^2 - 3^2 = -5 \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}$ So, the pair of numbers we got are $(3, 2)$ but 2 will have "$-$" sign as the question mentions that the imaginary part is negative. $\large \therefore \sqrt{{\color{#D61F06}5} - {\color{#3D99F6}12}i} = \pm ({\color{#20A900}3} - {\color{#20A900}2}i)$

Example 3: Find the value of $\large \sqrt{-{\color{#D61F06}5} + {\color{#3D99F6}12}i}$.

The procedure is same and the steps are same for every question. Here, in this example the imaginary part is positive but the real part is negative. So, we should choose the pair of numbers such that their difference will give the real part of given complex number. $\large \dfrac{\color{#3D99F6}12}{2} = {\color{#E81990}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies 3^2 - 2^2 = 5 \\ 2 \times 3 \implies \boxed{\color{#D61F06}2^2 - 3^2 = -5} \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}$ So, the pair of numbers we got are $(2, 3)$. As the imaginary part of the given complex number is positive the imaginary part of square root also have positive sign. $\large \therefore \sqrt{-{\color{#D61F06}5} + {\color{#3D99F6}12}i} = \pm ({\color{#20A900}2} + {\color{#20A900}3}i)$

Example 4: Find the value of $\large \sqrt{-{\color{#D61F06}5} - {\color{#3D99F6}12}i}$.

The procedure is same and the steps are same for every question. Here, in this example the imaginary part is positive but the real part is negative. So, we should choose the pair of numbers such that their difference will give the real part of given complex number. $\large \dfrac{\color{#3D99F6}12}{2} = {\color{#E81990}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies 3^2 - 2^2 = 5 \\ 2 \times 3 \implies \boxed{\color{#D61F06}2^2 - 3^2 = -5} \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}$ So, the pair of numbers we got are $(2, 3)$. As the imaginary part of the given complex number is negative the imaginary part of square root also have negative sign. $\large \therefore \sqrt{-{\color{#D61F06}5} - {\color{#3D99F6}12}i} = \pm ({\color{#20A900}2} - {\color{#20A900}3}i)$

PRACTICE PROBLEMS

• Find the square root of the following complex numbers : $\begin{array}{c}~1) 7 + 24i \qquad \quad 2) -7 - 24i \\ 3) 7 - 24i \qquad \quad 4) -7 + 24i \\ 5) 8 + 6i \qquad \quad 6) 8 - 6i \\ 7) -8 + 6i \qquad 8) -8 - 6i \\ 9) 3 + 4i \qquad \quad 10) 3 - 4i \\ 11) -3 + 4i \qquad 12) -3 - 4i \\ 13) i \qquad \qquad \quad 14) -i \\ \end{array}$

• Find all the possible values of $\sqrt{i} + \sqrt{-i}$

• Find the value of $\dfrac{\sqrt{3 + 4i} + \sqrt{3 - 4i}}{\sqrt{-3 + 4i} - \sqrt{-3 - 4i}}$

If you have any doubts or queries you can put them in the comments section. Also, you can keep your solutions for practice problems in the comments section.

$\large \color{teal} \text{THANK YOU !!!}$