Square Root of Complex Numbers (original) (raw)

Last Updated : 29 Jan, 2026

A complex number is a number that has two components, or is equally divided into a real part and an imaginary part.

Square-Root-of-Complex-Numbers

Square Root Formula for a Complex Number

Square Root of a Complex Number (Algebraic Method)

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:

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.

**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:

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.

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 = |z| = √a2 + b2 is the modulus of the complex number, and θ is the argument, given by:

θ = arg(z)

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

\sqrt{z} = \pm \sqrt{r}\left(\cos \frac{\theta}{2} + 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

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:

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)