How to use the Nth Root of Complex Number Calculator?

Here is the step-by-step guide to using the Nth root of a complex number calculator.

Step-1: Enter the Real Part:

Input the real part of the complex number into the “Real part” field. This is the numerical value associated with the real axis on the complex plane.

Step-2: Enter the Imaginary Part:

Input the imaginary part of the complex number into the “Imaginary part” field. This is the numerical value associated with the imaginary axis on the complex plane. Remember to enter ‘i’ after the value.

Step-3: Enter the Value of N:

Specify the value of ‘n’, which represents the root you want to calculate. For example, if you want to find the square root, enter ‘2’. For the cube root, enter ‘3’, and so on.

Step-4: Click “Calculate Nth Root”:

Once you’ve entered the real part, imaginary part, and value of ‘n’, click the “Calculate Nth Root” button.

Step-5: View the Results:

The calculator will display all possible roots of the complex number according to the specified ‘n’ value.

How to find the nth root of the complex number?

To find the nth root of a complex number, you can use De Moivre’s Theorem and the polar form of complex numbers. Here are the steps:

Step-1: Convert the complex number to polar form:

Write the complex number z=a+bi in polar form as z=r(cos(θ)+isin(θ)), where r is the magnitude of the complex number and θ is the angle of the complex number.

Step-2: Apply De Moivre’s Theorem:

The nth roots of a complex number can be found by raising the magnitude to the power of $\frac{1}{n}$​ and dividing the angle by n.

Step-3: Calculate the nth root:

After finding $r^{\frac{1}{n}}$ and $\frac{\theta }{n}$, the nth root is given by:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +360k}{n}\right)+i\sin \left(\frac{\theta +360k}{n}\right)\right)$

Where k=0,1,2,…,n-1.

NOTE: For the above formula, the angle should be in degrees.

We can also write in radian form,

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$

Where k=0,1,2,…,n-1.

NOTE: For the above formula, the angle should be in radians.

Example on the nth root of the complex number.

Find the square root of the complex number z=2+2i

Compare this complex number with z=x+iy

So here we have,

x=2, y=2

The magnitude is given by,

$r=\sqrt{x^2+y^2}\\$

$r= \sqrt{2^2+2^2}\\$

$r= \sqrt{4+4}\\$

$r= \sqrt{8}\\$

$r= 2\sqrt{2}\\$

The angle is given by,

$\theta =\tan ^{-1}\left(\frac{y}{x}\right)\\$

$\theta =\tan ^{-1}\left(\frac{2}{2}\right)\\$

$\theta =\tan ^{-1}\left(1\right)\\$

$\theta =\frac{\pi }{4}\\$

$\theta =45^{\circ }\\$

The trigonometric form of a complex number is given by,

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

The nth root of the complex number is given by,

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +360k}{n}\right)+i\sin \left(\frac{\theta +360k}{n}\right)\right)$

Here we have n=2.

$z^{\frac{1}{2}}=r^{\frac{1}{2}}\left(\cos \left(\frac{\theta \:+360k}{2}\right)+i\sin \left(\frac{\theta \:+360k}{2}\right)\right)\\$

Substitute all values.

$z^{\frac{1}{2}}=\left(2\sqrt{2}\right)^{\frac{1}{2}}\left(\cos \left(\frac{45+360k}{2}\right)+i\sin \left(\frac{45+360k}{2}\right)\right)\\$

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+360k}{2}\right)+i\sin \left(\frac{45+360k}{2}\right)\right)\\$

To find the first root, substitute k=0

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+360\cdot 0}{2}\right)+i\sin \left(\frac{45+360\cdot 0}{2}\right)\right)\\$

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+0}{2}\right)+i\sin \left(\frac{45+0}{2}\right)\right)\\$

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45}{2}\right)+i\sin \left(\frac{45}{2}\right)\right)\\$

$z^{\frac{1}{2}}=1.68179\left(0.92388+i0.38268\right)\\$

$z^{\frac{1}{2}}=1.5538a+0.6436i\\$

.

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+360k}{2}\right)+i\sin \left(\frac{45+360k}{2}\right)\right)\\$

To find the second root, substitute k=1

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+360\cdot 1}{2}\right)+i\sin \left(\frac{45+360\cdot 1}{2}\right)\right)\\$

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{45+360}{2}\right)+i\sin \left(\frac{45+360}{2}\right)\right)\\$

$z^{\frac{1}{2}}=2^{\frac{3}{4}}\left(\cos \left(\frac{405}{2}\right)+i\sin \left(\frac{405}{2}\right)\right)\\$

$z^{\frac{1}{2}}=1.68179\left(-0.92388-i0.38268\right)\\$

$z^{\frac{1}{2}}=-1.5538a-0.6436i\\$

The square root of the complex number z=2+2i are:

$First\:root:\:z_1=1.5538a+0.6436i\\$

$Second\:root:\:z_2=-1.5538a-0.6436i\\$

FAQs on the nth root of a complex number.

1) How do you find the nth root of a complex number?

Answer: To find the nth root of a complex number, convert it to polar form, then apply De Moivre’s Theorem by taking the nth root of its magnitude and dividing its argument by n.

2) How many nth roots does a complex number have?

Answer: A complex number has n distinct nth roots, where n is the degree of the root. These roots are evenly spaced around a circle in the complex plane.

3) What is the principal nth root of a complex number?

Answer: The principal nth root of a complex number is the root with the smallest argument, usually denoted as $z_0$. It’s commonly chosen as the one with the angle in the range (−𝜋,𝜋].

4) What is the geometric interpretation of nth roots in the complex plane?

Answer: The nth roots of a complex number are evenly spaced points around a circle in the complex plane, centered at the origin. The radius of this circle is the nth root of the magnitude of the original complex number.

5) How do you represent all nth roots of a complex number?

Answer: All nth roots of a complex number can be represented using the formula:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\\$

Where k=0,1,2,…,n-1.

NOTE: For the above formula, the angle should be in radians.

6) Can nth roots of a complex number be negative?

Answer: Yes, the nth roots of a complex number can have negative real parts or negative imaginary parts, or both, depending on the original complex number and the value of n.