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 \(\large \sqrt{{\color{red}5} + {\color{blue}12}i}\)

Step - 1 : Find the real part and imaginary part of the given complex number. \[\large \underbrace{\color{red}5}_{\text{Real Part}} \qquad + \underbrace{{\color{blue}12}}_{\text{Imaginary Part}}i\] Step - 2 : Take half of the co-efficient of the imaginary part.
\[\large \sqrt{5 + {\color{blue}12}i} \implies \dfrac{\color{blue}12}{2} = \color{pink}6\] Step - 3 : Express the obtained number as a product of two numbers. \[\large {\color{pink}6} = \begin{cases} 6 \times 1 \\ 3 \times 2 \\ 2 \times 3 \\ 1 \times 6 \\ \end{cases}\] 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. \[\large {\color{pink}6} = \begin{cases} 6^2 - 1^2 = 35 \\ \boxed{\color{red}3^2 - 2^2 = 5} \\ 2^2 - 3^2 = -5 \\ 1^2 - 6^2 = -35 \\ \end{cases}\] 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. \[\large \therefore \sqrt{{\color{red}5} + {\color{blue}12i}} = \pm ({\color{green}3} + {\color{green}2}i)\]


Example 2 : Find the value of \(\large \sqrt{{\color{red}5} - {\color{blue}12}i}\).

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. \[\large \dfrac{\color{blue}12}{2} = {\color{pink}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies \boxed{\color{red}3^2 - 2^2 = 5} \\ 2 \times 3 \implies 2^2 - 3^2 = -5 \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}\] 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. \[\large \therefore \sqrt{{\color{red}5} - {\color{blue}12}i} = \pm ({\color{green}3} - {\color{green}2}i)\]


Example 3: Find the value of \(\large \sqrt{-{\color{red}5} + {\color{blue}12}i}\).

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. \[\large \dfrac{\color{blue}12}{2} = {\color{pink}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies 3^2 - 2^2 = 5 \\ 2 \times 3 \implies \boxed{\color{red}2^2 - 3^2 = -5} \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}\] 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. \[\large \therefore \sqrt{-{\color{red}5} + {\color{blue}12}i} = \pm ({\color{green}2} + {\color{green}3}i)\]


Example 4: Find the value of \(\large \sqrt{-{\color{red}5} - {\color{blue}12}i}\).

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. \[\large \dfrac{\color{blue}12}{2} = {\color{pink}6} = \begin{cases} 6 \times 1 \implies 6^2 - 1^2 = 35 \\ 3 \times 2 \implies 3^2 - 2^2 = 5 \\ 2 \times 3 \implies \boxed{\color{red}2^2 - 3^2 = -5} \\ 1 \times 6 \implies 1^2 - 6^2 = -35 \\ \end{cases}\] 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. \[\large \therefore \sqrt{-{\color{red}5} - {\color{blue}12}i} = \pm ({\color{green}2} - {\color{green}3}i)\]


PRACTICE PROBLEMS

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

  • 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.


Want to improve your mathematics !!! Try my set Mathematics Done Right

\[\large \color{teal} \text{THANK YOU !!!}\]

Note by Ram Mohith
2 weeks, 3 days ago

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You really have some good presentation skills.

Abha Vishwakarma - 1 week, 5 days ago

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Thanks. But how is the tequnic.

Ram Mohith - 1 week, 5 days ago

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Quite useful. But is it good for larger values of the real and imaginary parts? Because larger numbers would take time to factorize. And squaring the factors would be an issue if the factors themselves get very large.

I think its very useful for most of the calculations since really large numbers are not involved in most cases. But I would not go for it if the numbers get very large because I do mistakes in big calculations.

Abha Vishwakarma - 1 week, 5 days ago

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@Abha Vishwakarma Going by formula will also be difficult if large numbers are given. But as you told such large numbers will not be asked anywhere.

Ram Mohith - 1 week, 5 days ago

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quires is actually spelt queries.

Mohmmad Farhan - 5 days, 5 hours ago

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There's too much trial and error involved in this "trick". Why not just simply solve it algebraically?

In the example you gave (\(5+12i\)), if a complex number \( z \equiv a + bi \) solves \( z^2 = 5 + 12i \), then we must have that \(a^2 - b^2 = 5\) and \(2ab = 12\), for which the latter gives us \(ab = 6\). Substituting this into the former result and simplifying as a polynomial in, say, \(a\), we get \(a^4- 5a^2 - 36 = 0 \). Factorise the left-hand side to get \( (a^2+4)(a-3)(a+3) = 0 \). As the components of \(z\) cannot be complex numbers by prescription, the factor \( a^2 + 4 \) gives extraneous solutions, and hence we have that \(a = \pm 3\), for which we obtain \(b = 2\) when \(a = 3\) and \(b = -2\) when \(a = -3\). Hence, \( z = \pm (3 + 2i) \) solves \(z^2 = 5 + 12i\).

My point in presenting that is that it is far less expedient to solve algebraically than resort to "guessing" tricks whenever possible. Even in this case, there is also no need to resort to remembering formulas, but rather methodologies (which is an essential skill for any mathematician).

Gennady Notowidigdo - 1 week, 2 days ago

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This is not trail and error. And for your confirmation this trick is very much usefull in objective exams where you must solve more questions in limited time. So, this trick is a great help.

Ram Mohith - 1 week, 2 days ago

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Substituting factor pairs and checking which ones satisfy an equation is the very definition of “trial and error”. Works well when you have a small amount of checks, but complexity is increased exponentially (logarithmically, to be precise) as the numbers get larger. You must consider the computational implications of your method with respect to the other methods out there, and ultimately the algebraic one I have presented is the quickest one. This is the method used in most universities these days, and the appeal in this method is in its precision relative to the time taken.

Gennady Notowidigdo - 1 week, 2 days ago

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@Gennady Notowidigdo Whatever you say, this method is very useful for competitive exams under university level.

Ram Mohith - 1 week, 2 days ago

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