Square root of complex numbers

Complex Number

A complex number is a type of number that includes both real and imaginary parts. It is used to represent quantities that cannot be described by real numbers alone. The general form of a complex number is written as $a + ib$, where $i$ is the imaginary unit and $i = \sqrt{-1}$.

A number of the form $a + ib$ is called a complex number, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit. We usually denote a complex number by letters like $z, z_1, z_2$, etc.

Example:
For example, $z = 5 + 2i$ is a complex number.
Here, $5$ is the real part, denoted by $\operatorname{Re}(z)$, and $2$ is the imaginary part, denoted by $\operatorname{Im}(z)$.

What is the Square Root of a Complex Number?

The square root of a complex number is another complex number whose square is equal to the original complex number. If $\sqrt{a + ib} = x + iy$, then squaring both sides gives $(x + iy)^2 = a + ib$. By comparing the real and imaginary parts, we can find the values of $x$ and $y$.

This method is useful in solving quadratic equations and in various mathematical applications.

Formula for Finding Square Roots of Complex Numbers

The square root of a complex number can be calculated using a few key formulas. These formulas are derived by comparing real and imaginary parts or by using polar coordinates and Euler’s formula. Below are the important formulas and their applications.

General Formula for $\sqrt{a + bi}$

For a complex number $z = a + bi$, its square root can be written as:

$\sqrt{a + bi} = \pm \left(\sqrt{\frac{|z| + a}{2}} + i \, \text{sgn}(b) \sqrt{\frac{|z| - a}{2}}\right)$

Where:

• $|z| = \sqrt{a^2 + b^2}$ is the modulus of $z$

• $\text{sgn}(b)$ is the sign of $b$, which determines the sign of the imaginary part

This formula gives both solutions depending on the $\pm$ sign. It’s particularly useful when solving problems in Cartesian form.

Expressing Square Roots in Polar Form

If the complex number is given in polar form as: $z = r(\cos \theta + i \sin \theta)$

Then its square roots are expressed as: $\sqrt{z} = \sqrt{r} \left[\cos \frac{\theta + 2k\pi}{2} + i \sin \frac{\theta + 2k\pi}{2}\right], \quad k = 0, 1$

This formula shows that the modulus of the square root is the square root of $r$, and the argument is half of $\theta$ plus $k\pi$ to account for both solutions.

Principal and Secondary Square Roots

A complex number always has two square roots:

1. Principal Square Root
   The root where $k = 0$ is called the principal square root. It is commonly used in problems and has the smallest argument.

2. Secondary Square Root
   The second root is found by setting $k = 1$. This root differs by $180^\circ$ from the principal root in the Argand plane.

Together, these two roots satisfy the equation $w^2 = z$, where both $w$ and $-w$ are valid solutions.

How to Find the Square Root of Complex Numbers?

Let $z = x + iy$ be the complex number whose square root we need to find. Since the square root must also be a complex number, we assume:

$z^{1/2} = a + ib$

Squaring Both Sides

$z = x + iy = (a + ib)^2 = a^2 - b^2 + 2iab$

Comparing Real and Imaginary Parts

$\begin{aligned} a^2 - b^2 &= x \\ 2ab &= y \\ a^2 + b^2 &= \sqrt{x^2 + y^2} = |z| \end{aligned}$

Solving for $a$ and $b$

From the above equations, we can derive:

$a = \pm \sqrt{\frac{x + |z|}{2}}$

$b = \pm \sqrt{\frac{|z| - x}{2}}$

Thus, the square root of $z$ is given by:

$\sqrt{z} = \pm \left( \sqrt{\frac{|z| + \operatorname{Re}(z)}{2}} + i \sqrt{\frac{|z| - \operatorname{Re}(z)}{2}} \right)$

Note: If $\operatorname{Im}(z) > 0$, the formula uses a positive sign between the real and imaginary parts; otherwise, a negative sign is used.

Square Root of Complex Numbers in Polar Form

In this section, we explore how to find the square root of complex numbers using polar coordinates and the nth root formula. This method simplifies calculations by working with the modulus and argument of the complex number.

Using the nth Root Theorem

The square root of a complex number can also be found using the $n^{th}$ root theorem in polar form. If the complex number is written as: $z = r(\cos \theta + i \sin \theta)$

Then, the $n^{th}$ roots are given by: $z^{1/n} = r^{1/n} \left[\cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n}\right]$

For square roots, i.e., $n=2$, the formula becomes: $z^{1/2} = \sqrt{r} \left[\cos \frac{\theta + 2k\pi}{2} + i \sin \frac{\theta + 2k\pi}{2}\right]$ where $k = 0, 1$.

Important Note: If one square root is $a + ib$, the other must be its negative $-(a + ib)$.

Step-by-Step Method to Find Square Roots of Complex Numbers

Finding the square root of a complex number involves both mathematical formulas and a clear understanding of its properties. Below are methods using Cartesian, polar, and Euler forms to calculate these roots.

Solving Using Cartesian Form

A complex number in Cartesian form is written as $z = x + iy$. Its square root can be assumed as another complex number:

$\sqrt{z} = a + ib$

Squaring both sides:

$(a + ib)^2 = a^2 - b^2 + 2abi = x + iy$

Now, by comparing real and imaginary parts:

$a^2 - b^2 = x$
$2ab = y$

We also use the modulus:

$a^2 + b^2 = |z| = \sqrt{x^2 + y^2}$

Solving for $a$ and $b$ gives:

$a = \pm \sqrt{\frac{x + |z|}{2}}$
$b = \pm \sqrt{\frac{|z| - x}{2}}$

These formulas allow you to find both roots depending on the sign chosen.

Solving Using Polar Coordinates

In polar form, the complex number is:

$z = r(\cos \theta + i \sin \theta)$

Its square root is:

$\sqrt{z} = \sqrt{r}\left[\cos \frac{\theta + 2k\pi}{2} + i \sin \frac{\theta + 2k\pi}{2}\right], \quad k = 0, 1$

This formula helps find both roots using the argument $\theta$ and modulus $r$.

Using Euler’s Formula to Find Square Roots

Euler’s formula expresses complex numbers as: $z = r e^{i\theta}$

Thus, the square root becomes: $\sqrt{z} = \sqrt{r} e^{i\frac{\theta}{2}}$

For both solutions: $\sqrt{z} = \sqrt{r} e^{i(\frac{\theta}{2} + k\pi)}, \quad k = 0, 1$

This is a compact way to calculate roots using exponential notation.

Worked Examples of Square Roots of Complex Numbers

Here we solve practical problems to find the square roots of complex numbers using formulas in Cartesian and polar forms. These examples help understand how to apply the square root of complex numbers in real calculations.

Example 1 – $\sqrt{3 + 4i}$

Modulus: $|z| = \sqrt{3^2 + 4^2} = 5$

Argument: $\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$

Square root using polar form:

$\sqrt{z} = \sqrt{5}\left[\cos \frac{53.13^\circ}{2} + i \sin \frac{53.13^\circ}{2}\right]$

$= \sqrt{5} (\cos 26.565^\circ + i \sin 26.565^\circ)$

Approximate values:

$\approx 2(0.8944 + i 0.4472)$
$\approx 1.7888 + i 0.8944$

The second root is obtained by setting $k=1$.

Example 2 – $\sqrt{-5 + 12i}$

Modulus: $|z| = \sqrt{(-5)^2 + 12^2} = 13$

Argument: $\theta = \tan^{-1}\left(\frac{12}{-5}\right)$

Since it's in the second quadrant, $\theta \approx 112.62^\circ$.

Square root: $\sqrt{z} = \sqrt{13} \left[\cos \frac{112.62^\circ}{2} + i \sin \frac{112.62^\circ}{2}\right]$

$= \sqrt{13} (\cos 56.31^\circ + i \sin 56.31^\circ)$

Approximate values can be computed accordingly.

Example 3 – Finding Both Solutions of $\sqrt{1 + i}$

Modulus: $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$

Argument: $\theta = \tan^{-1}\left(\frac{1}{1}\right) = 45^\circ$

Square root: $\sqrt{z} = \sqrt[4]{2}\left[\cos \frac{45^\circ}{2} + i \sin \frac{45^\circ}{2}\right]$

For $k = 0$: $= \sqrt[4]{2}\left(\cos 22.5^\circ + i \sin 22.5^\circ\right)$

For $k = 1$: $= \sqrt[4]{2}\left(\cos 202.5^\circ + i \sin 202.5^\circ\right)$

This illustrates both roots geometrically.

Properties of Square Roots of Complex Numbers

The square roots of complex numbers follow certain important rules that help in understanding their behavior in the complex plane. Below are the main properties that make them useful in solving mathematical problems

• Always Two Solutions – Every complex number has exactly two square roots, which are negatives of each other. If one root is $a + bi$, the other is $-(a + bi)$.

• Relationship with Modulus – The modulus of each square root is equal to $\sqrt{|z|}$, where $|z| = \sqrt{a^2 + b^2}$.

• Relationship with Argument – The argument of each square root is half of the argument of the original complex number. If $\arg(z) = \theta$, then the square roots have arguments $\frac{\theta}{2}$ and $\frac{\theta}{2} + \pi$.

• Symmetry in the Complex Plane – The two roots are symmetric with respect to the origin and lie on opposite sides along a line passing through the origin at an angle related to $\frac{\theta}{2}$.

• Expressed in Polar Form – The square roots can be written as $\sqrt{r} \left[ \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right]$ and $\sqrt{r} \left[ \cos\left(\frac{\theta}{2} + \pi\right) + i \sin\left(\frac{\theta}{2} + \pi\right) \right]$.

Solved Examples Based on Square Root of Complex Numbers

Example 1: If $(x+i y)^2=7+24 i$ then a value of $(7+\sqrt{-576})^{1 / 2}-(7-\sqrt{-576})^{1 / 2}$ is:

Solution: As we have learned in,

Square Root of a Complex Number - $\sqrt{z}=a+i b \text { where } z=x+i y$ is calculated by equating real and imaginary parts of $x+i y=(a+i b)^2$

We have to find $(7+24 i)^{1 / 2}-(7-24 i)^{1 / 2}$

Now $(x+i y)^2=7+24 i$

$x+i y=\sqrt{7+24 i}$

and $x-i y=\sqrt{7-24 i}$ (Put - i at the place of i)

$\therefore \quad i 2 y=\sqrt{7+24 i}-\sqrt{7-24 i}$

Now $x^2-y^2+i 2 x y=7+24$

$\therefore \quad x^2-y^2=7, x y=12$

Solving we get $x= \pm 4 ; y \pm 3$,

$\therefore \quad i 2 y= \pm 6 i$

$\therefore-6i$

Hence, the answer is -6i.

Example 2: One of the square roots of $21-20 i$ equals.

Solution:

Let a + ib be the square root

So, $(a+i b)^2=21-20 i$

$a^2-b^2+i(2 a b)=21-20 i$

Thus,

$a^2-b^2=21 \text { and } 2 a b=-20$

$a^2-b^2=21 \text { and } a h=-10$

Using the relation $\left(a^2+b^2\right)^2=\left(a^2-b^2\right)^2+4 a^2 b^2$
$\left(a^2+b^2\right)^2=21^2+20^2=841$
$a^2+b^2=29$

Adding this equation to $\mathrm{a}^2-\mathrm{b}^2=21$, we get

$a^2=25$

$\therefore a=5$ or $a=-5$

Now using ab = -10, when a= 5 then b= -2 and when a =-5, b= 2

$a+i b=5-2 i$ or $-5+2 i$

Hence, the answer is -5+2i.

Example 3: If $z=x-i y$ and $z^{\frac{1}{3}}=p+i q$ then $\frac{\left(\frac{x}{p}+\frac{q}{q}\right)}{\left(p^2+q^2\right)}$ is equal to

Solution:

Equality in Complex Numbers -

$z=x+i y \& w=a+i b$ are equal if $x=a \& y=b$

where in

Two complex numbers are equal if real parts as well as imaginary parts are equal.

$p=x-i y \quad$ and $z^{1 / 3}=p+i d$

$\therefore \quad z=(p+i q)^3$

$\therefore x-i y=p^3+(i q)^3+3 p \cdot i q(p+i q)=p^3+i^3 q^3+i p^2 q-3 p q^2$

$x-i y=\left(p^3-3 p q^2\right)+i\left(3 p^2 q-q^3\right)$

Compare: $p^3-3 p q^2=x$

$p\left(p^2-3 q^2\right)=x$

$p^2-3 q^2=\frac{x}{p}$

and $q\left(3 p^2-q^2\right)=-y$

$q^2-3 p^2=\frac{y}{q}$

$\therefore \frac{\frac{x}{p}+\frac{y}{q}}{p^2+q^2}=\frac{p^2-3 q^2+q^2-3 p^2}{\left(p^2+q^2\right)}=\frac{-2\left(p^2+q^2\right)}{p^2+q^2}=-2$

Hence, the answer is -1.

List of topics related to the Square Root of a Complex Number

The square root of a complex number is connected to several related concepts that help in building a deeper understanding of the topic. Below is a list of important topics associated with it.

NCERT Resources

NCERT resources make it easy to learn and practice complex numbers and quadratic equations by providing detailed solutions, exemplars, and handy notes. The following links are essential for building a strong foundation in this topic, and help in both conceptual clarity and exam preparation.

NCERT Solutions for Class 11 Maths Chapter 5 -Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Class 11 Maths Notes for Chapter 5 - Complex Numbers and Quadratic Equations

Practice Questions based on the Square Root of a Complex Number

Below are practice questions based on the square root of complex numbers to enhance problem-solving skills and prepare for exams. These help develop a clear understanding of the methods used to find square roots in complex numbers.

Square Root Of Complex Numbers - Practice Question MCQ

We have shared practice question sets for related topics below: