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Square Root of Complex Numbers

Last Updated : 17 Oct, 2024
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A complex number is a number that has two components: or equally divided into a real part and an imaginary part. The general form of a complex number is:

z = a + bi

where:

  • a is the real part.
  • b is the imaginary part, and i is the imaginary unit, where i 2 = -1.
Square-Root-of-Complex-Numbers
Square Root of Complex Numbers

Square Root of Complex Numbers Calculator

You can use the following calculator to calculate the square root of any complex number:

Square Root of a Complex Number

To find the t of a complex number z = a + bi, we need to find another complex number w = x + yi such that:

w2 = z

This means: (x + yi)2 = a + bi

x2 + 2xyi - y2 = a + bi

Equating real and imaginary parts:

  • x 2 - y 2 = a (Real part)
  • 2xy = b (Imaginary part)

Now, Solve for x and y:

First, let's solve for x2 and y2. To do this, substitute y = \frac{b}{2x} into the first equation (from 2xy = b):

x^2 - \left( \frac{b}{2x} \right)^2 = a

x^2 - \frac{b^2}{4x^2} = a

4x^4 - b^2 = 4a x^2

4x^4 - 4a x^2 - b^2 = 0

Let z = x^2 to simplify this quadratic equation:

4z^2 - 4a z - b^2 = 0

Solve this quadratic equation using the quadratic formula:

z = \frac{-(-4a) \pm \sqrt{(-4a)^2 - 4(4)(-b^2)}}{2(4)}

z = \frac{4a \pm \sqrt{16a^2 + 16b^2}}{8}

z = \frac{4a \pm 4\sqrt{a^2 + b^2}}{8}

z = \frac{a \pm \sqrt{a^2 + b^2}}{2}

Since z = x^2, we have:

x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}

Thus, x =\pm \sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}

Now use y = b/2x to find y:

y = \pm\frac{b}{2 \sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}}

y =\pm \frac{b}{\sqrt{2}} \times \frac{1}{\sqrt{\frac{\sqrt{a^2 + b^2} + a}{\sqrt{a^2 + b^2} - a} \times \sqrt{a^2 + b^2 - a}}}

y =\pm \frac{b}{\sqrt{2}} \times \frac{\sqrt{\sqrt{a^2 + b^2} - a}}{\sqrt{a^2 + b^2 - a^2}}

y = \pm\frac{b}{\sqrt{2}} \times \frac{\sqrt{\sqrt{a^2 + b^2} - a}}{\sqrt{b^2}}

y =\pm \frac{b}{|b|} \times \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}

Result: The square root of a + bi is given by:

x + iy = \pm \left(\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}} + i \cdot \frac{b}{|b|} \times \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\right)

As we know, |z| = \sqrt{a^2 + b^2} ,

Thus, \sqrt{z} = x + iy = \pm \left( \sqrt{\frac{a + |z|}{2}} + i \cdot \frac{b}{|b|} \sqrt{\frac{|z| - a}{2}}\right)

This is the general form for the square root of a complex number a + bi.

Let's consider an example for better understanding.

Example: Find square root of 4 + 3i.

Solution:

Let z = \sqrt{4 + 3i} = x + iy, which implies:

z^2 = (x + iy)^2 = x^2 + 2ixy - y^2

Equating the real and imaginary parts with 4 + 3i, we get:

  • x2 - y2 = 4 (Real part)
  • 2xy = 3 (Imaginary part)

From 2xy = 3, solve for y:

y = 3/2x

Substitute this into x2 - y2 = 4:

x^2 - \left( \frac{3}{2x} \right)^2 = 4

\Rightarrow x^2 - \frac{9}{4x^2} = 4

⇒ 4x4 - 9 = 16x2

⇒ 4x4 - 16x2 - 9 = 0

Let u = x2, so the equation becomes:

⇒ 4u2 - 16u - 9 = 0

Solve this quadratic using the quadratic formula:

\Rightarrow u = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(4)(-9)}}{2(4)}

\Rightarrow u = \frac{16 \pm \sqrt{256 + 144}}{8} = \frac{16 \pm \sqrt{400}}{8} = \frac{16 \pm 20}{8}

\Rightarrow u = \frac{16 + 20}{8} = \frac{36}{8} = 9/2 or u = \frac{16 - 20}{8} = \frac{-4}{8} = -1/2

Since u = x2 ≥ 0, we discard u = -1/2.

Thus, x2 = 9/2, so x = \pm \sqrt{\frac{9}{2}} = \pm \frac{3}{\sqrt{2}}

Now, substitute x into y = 3/2x:

\Rightarrow y =\pm \frac{3}{2 \times \frac{3}{\sqrt{2}}} = \pm \frac{1}{\sqrt{2}}

Thus, z = x + iy =\pm \frac{3}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}}.

\Rightarrow \sqrt z = x + iy = \pm \left(\frac{3}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}} \right).

Note: Alternatively we can use the provided formula for easy calculations.

How to Find Square Root of a Complex Number (Polar Form)?

The following procedure can also be followed to find the square root of a complex number a + bi, when given in polar form.

Step 1: Express the complex number in polar form. The polar form of a complex number z = a + bi is written as:

z = r(cos θ + isin θ)

Where r = a2 + b2 is the modulus of the complex number, and θ is the argument, given by:

θ = tan - 1(b / a)

Step 2: Using the square root formula from polar form. The square root of a complex number in polar form is:

z = r \left( \cos \frac{\theta}{2} + \mathbf{i} \sin \frac{\theta}{2} \right)

Step 3: Calculate the modulus r and argument θ, then by the help of below formula, find the square root of complex number.

Note: Prefer this method only if given complex number is z = r(cos θ + isin θ) = reiθ, and θ is well known value such as π/2, π/4, π/3, π/6, etc.

Let's consider an example for better understanding:

Example: Find the square root of z = eiπ/3.

Solution:

Given: z = eiπ/3 = z = cos π/3 + isin π/3

  • r =1
  • θ = π/3

Use the formula,

\sqrt z = \left( \cos \frac{\pi/3}{2} + i \sin \frac{\pi/3}{2} \right)
\Rightarrow \sqrt z = \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)
\Rightarrow \sqrt z = \left( \frac{\sqrt 3}{2} + i \frac{1}{2} \right)

General Formula for nth Root of Complex Number

Given a complex number z = r(cos⁡θ + isin⁡θ), the nth roots of z are:

z_k = \sqrt[n]{r} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right)

Where:

  • r is the modulus of z i.e., r = ∣z∣
  • θ is the argument (angle) of the complex number,
  • n is the root you want to find,
  • k is an integer and takes values from 0 to n−1.

Solved Problems on Square Root of Complex Numbers

Problem 1: Find square root of z = 3 + 4i

Solution:

|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Apply the formula: \sqrt{z} = \pm \left( \sqrt{\frac{3 + 5}{2}} + i \cdot \frac{4}{|4|} \sqrt{\frac{5 - 3}{2}} \right)
\Rightarrow \sqrt{z} = \pm \left( \sqrt{\frac{8}{2}} + i \cdot 1 \cdot \sqrt{\frac{2}{2}} \right)
\Rightarrow \sqrt{z} = \pm \left( \sqrt{4} + i \sqrt{1} \right) = \pm (2 + i)

Problem 2: Find square root of z = 1 + i.

Solution:

|z| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}

Apply the formula: \sqrt{z} = \pm \left( \sqrt{\frac{1 + \sqrt{2}}{2}} + i \cdot \frac{1}{|1|} \sqrt{\frac{\sqrt{2} - 1}{2}} \right)​

After calculating, we get: \sqrt{z} = \pm \left( \sqrt{1.207} + i \sqrt{0.207} \right) \approx \pm (1.099 + 0.455i)

Problem 3: Find square root of z = − 1 + i.

Solution:

|z| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}

Apply the formula: \sqrt{z} = \pm \left( \sqrt{\frac{-1 + \sqrt{2}}{2}} + i \cdot \frac{1}{|1|} \sqrt{\frac{\sqrt{2} + 1}{2}} \right)

After calculating, we get: \sqrt{z} = \pm \left( \sqrt{0.207} + i \sqrt{1.207} \right) \approx \pm (0.455 + 1.099i)

Problem 4: Find square root of z = − 4 + 0i (Purely Real).

Solution:

|z| = \sqrt{(-4)^2} = \sqrt{16} = 4

Apply the formula: \sqrt{z} = \pm \left( \sqrt{\frac{-4 + 4}{2}} + i \cdot \frac{0}{|0|} \sqrt{\frac{4 - (-4)}{2}} \right)
\Rightarrow \sqrt{z} = \pm \left( \sqrt{0} + i \sqrt{4} \right) = \pm (0 + 2i)

Hence, −4 = ±2i

Problem 5: Find square root of z = −3 + 4i.

Solution:

|z| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Apply the formula: \sqrt{z} = \pm \left( \sqrt{\frac{-3 + 5}{2}} + i \cdot \frac{4}{|4|} \sqrt{\frac{5 + 3}{2}} \right)
\Rightarrow \sqrt{z} = \pm \left( \sqrt{1} + i \sqrt{4} \right) = \pm (1 + 2i)

Hence, −3 + 4i = ±(1 + 2i)

Worksheet: Square Root of Complex Numbers

Worksheet-on-Square-Root-of-Complex-Numbers

You can download this free worksheet on Square Root of Complex Numbers from below:

Download Free Worksheet on Square Root of Complex Numbers

Conclusion

In conclusion, finding the square root of complex numbers may seem challenging at first, but once you understand the process, it becomes much clearer. By expressing a complex number in its polar form and applying basic mathematical operations, you can calculate its square root.

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