This calculator finds the square root, modulus, inverse, conjugate and converts complex numbers to polar form. The solver explains each operation in detail, step by step.
This calculator performs five operations on a single complex number.
It computes module, conjugate, inverse, roots and polar form.
The modulus or magnitude of a complex number ( denoted by $ \color{blue}{ | z | }$ ), is the distance between the origin and that number.
If the $ z = a + bi $ is a complex number than the modulus is
$$ |z| = \sqrt{a^2 + b^2} $$Example 01: Find the modulus of $ z = \color{blue}{6} + \color{purple}3{} i $.
In this example $ \color{blue}{a = 6} $ and $ \color{purple}{b = 3} $, so the modulus is:
$$ \begin{aligned} | z | &= \sqrt{ a^2 + b^2} = \sqrt{6^2 + 3^2} = \\[1 em] &= \sqrt{36 + 9} = \sqrt{45} = \\[1 em] &= \sqrt{9 \cdot 5} = 3 \sqrt{5} \end{aligned} $$To find the complex conjugate of a complex number, we need to change the sign of the imaginary part. The conjugate of $ z = a \color{red}{ + b}\,i $ is:
$$ \overline{z} = a \color{red}{ - b}\,i $$Example 02: The complex conjugate of $~ z = 3 \color{blue}{+} 4i ~$ is $~ \overline{z} = 3 \color{red}{-} 4i $.
Example 03: The conjugate of $~ z = - 4i ~$ is $~ \overline{z} = 4i $.
Example 04: The conjugate of $~ z = 15 ~$ is $~ \overline{z} = 15 ~$, too.
The inverse or reciprocal of a complex number $ a + b\,i $ is
$$ \color{blue}{ \frac{1}{a + b\,i}} $$Here is an example of how to find the inverse.
Example 06: Find the inverse of the number $ z = 4 + 3i $
$$ \begin{aligned} \frac{1}{z} &= \frac{1}{4+3i} = \frac{1}{4+3i} \cdot \frac{4-3i}{4-3i} = \\[1 em] &= \frac{4-3i}{(4+3i)(4-3i)} = \frac{4-3i}{4^2 - (3i)^2} = \\[1 em] &= \frac{4-3i}{16+9} = \frac{4-3i}{25} = \frac{4}{25} - \frac{3}{25} i \end{aligned} $$The polar form of a complex number $ z = a + i\,b$ is given as $ z = |z| ( \cos \alpha + i \sin \alpha) $.
Example 05: Express the complex number $ z = 2 + i $ in polar form.
To find a polar form, we need to calculate $|z|$ and $ \alpha $ using formulas in the above image.
$$ |z| = \sqrt{2^2 + 1^2} = \sqrt{5}$$ $$ \tan \alpha = \dfrac{b}{a} = \dfrac{1}{2} \implies \alpha = \tan^{-1}\left(\dfrac{1}{2}\right) \approx 27^{o} $$So, the polar form is:
$$ z = \sqrt{5} \left( \cos 27^{o} + i \sin27^{o} \right) $$Please tell me how can I make this better.