**Roots of Complex Numbers**

What is the square root of a complex number? Complex numbers also have two square roots; the principal square root of a complex number z, denoted by sqrt(z), is always the one of the two square roots of z with a positive imaginary part. The square roots of *a* + *bi* (with *b* ≠ 0) are , where

and

where sgn is the signum function. This can be seen by squaring to obtain *a* + *bi*.

^{}Here **√a²+b²** is called the modulus of *a* + *bi*, and the square root sign indicates the square root with non-negative real part, called the **principal square root**; also where *z* = *a* + *bi*.

**Principal square root of a complex number**

To find a definition for the square root that allows us to consistently choose a single value, called the principal value, we start by observing that any complex number *x* + *iy* can be viewed as a point in the plane, (*x*, *y*), expressed using Cartesian coordinates.

The same point may be reinterpreted using polar coordinates as the pair ), where *r* ≥ 0 is the distance of the point from the origin, and varphi is the angle that the line from the origin to the point makes with the positive real (*x*) axis. In complex analysis, the location of this point is conventionally written If

then we define the principal square root of *z* as follows:

The principal square root function is thus defined using the nonpositive real axis as a branch cut. The principal square root function is holomorphic everywhere except on the set of non-positive real numbers (on strictly negative reals it isn’t even continuous). The above Taylor series for **√1+x** remains valid for complex numbers *x* with |*x*| < 1.

The above can also be expressed in terms of trigonometric functions:

**Formula Roots of Complex Numbers**

**De Moivre’s Theorem: **Let *z* = *r*(cos(*θ*) + ısin(*θ*).*Thenz*^{n} = [*r*(cos(*θ*) + ısin(*θ*)]^{n} = *r*^{n}(cos(*nθ*) + ısin(*nθ*), where *n* is any positive integer.

**Roots Of A Complex Number: **A complex number *z* = *r*(cos(*θ*) + ısin(*θ*) has exactly *n**n*th roots given by the equation *cos*(*n* is a positive integer, and *k* = 0, 1, 2,…, *n* – 2, *n* – 1.

**How do you find the square roots of a complex number?**

Every complex number has complex square roots. However since we don’t know how to deal with expressions such as √i we need to follow a specific method to find the square roots of a complex number.

**Let’s consider the complex number 21-20i.**

We know that all square roots of this number will satisfy the equation 21-20i=x^{2} by definition of a square root.

We also know that x can be expressed as a+bi (where a and b are real) since the square roots of a complex number are always complex.

**So 21-20i=(a+bi) ^{2}.**

The natural step to take here is the mulitply out the term on the right-hand side.

This gives 21-20i=a^{2}+(2ab)i+(b^{2})i^{2}.

As i^{2}=-1 by definition of i, this equation can be rearranged to give 21-20i=(a^{2}-b^{2})+(2ab)i.

Now both sides of the equation are in the same form.

**Let’s compare coeffiecients to obtain two equations in a and b.**

First, let’s compare the real parts of the equation.

We have a^{2}-b^{2}=21 (call this equation 1).

Next, let’s compare the imaginary parts of the equation (the coefficients of i).

We have 2ab=-20 (call this equation 2).

**We now have two equations in two unknowns. We can solve these simultaneous equations for a and b.**

Firstly, we can make b the subject of equation 2 by dividing both sides by 2a.

We have b=-10/a.

Now substitute this expression for b into equation 1.

We have a^{2}-(-10/a)^{2}=21.

Some simplification and factorisation of this equation gives us (a^{2}+4)(a^{2}-25)=0, a quadratic in disguise.

So either a^{2}=-4 or a^{2}=25.

We have assumed a to be real so a^{2}=-4 has no solutions of interest to us.

This means our solutions are a=5 and a=-5.

Substitute each a value into our earlier expression for b.

**This means that when a=5, b=-2 and when a=-5, b=2.**

**So, putting a and b back into the context of the question, we have two solutions: 5-2i and -5+2i.**

**How to use this complex root calculator?**

To use the complex root calculator, you can follow these steps:

- Enter the form of the complex number which you want to determine the complex roots of: you can choose between inputting it in its
`Cartesian form`

or its`polar form`

. - Tell us the degree
`n`

of the root you are interested in. - Our complex root calculator returns all
`n`

-th roots of the number you entered. - You can choose the form in which the calculator displays the results (the Cartesian form or the polar form).
- If you need it, go to the
`advanced mode`

to increase the`precision`

(number of decimal places) with which the complex root calculator displays its results. By default, we use`4`

decimal places.

**How to find complex roots by hand?**

As for finding roots of complex numbers by hand, you can choose either an algebraic or geometric way. You can find both approaches below.

To algebraically find the `n`

-th complex roots of a complex number `z`

, follow these steps:

- If your number
`z`

is given as its Cartesian coordinates,`a + bi`

, convert it to the**polar form**. In other words, find its magnitude`r`

and argument`φ`

. - Compute the
`n`

-th root of`r`

. - Compute
`φ / n`

and its multiplicities:`2 * φ / n`

,`3 * φ / n`

, up to`(n-1) * φ / n`

. - You can find the roots you’re looking for using the following formula
`ⁿ√r * exp(i * (θ + 2kπ) / n)`

or, equivalently,`ⁿ√r * (cos(θ/n + 2kπ/n) + i * sin(θ/n + 2kπ/n))`

where

`k = 0, ..., n - 1`

.

To geometrically find all `n`

-th roots of `z`

, follow these steps:

- The first step is the same as in the algebraic approach: if your number is in its Cartesian form, compute its magnitude
`r`

and argument`φ`

. **Plot a circle**with a radius equal to the`n`

-th root of`r`

. All the roots we’re searching for lie in this circle.- On this circle,
**mark a point with argument**. This number is one of the roots and the starting point to find all the remaining ones!`φ / n`

- Starting from this first root,
**make marks at every**until you make a full circle and come back to the starting point. These points are all the other complex roots you’re looking for.`2π / n`

radians

**Roots of Complex Numbers Exercises**

**1. Find the square root of 8 – 6i:**

Answer:

Let z² = (x + yi)² = 8 – 6i

\ (x² – y²) + 2xyi = 8 – 6i

Compare real parts and imaginary parts,

x² – y² = 8 (1)

2xy = -6 (2)

Now, consider the modulus: |z|² = |z²|

\ x² + y² = √(8² + 6²) = 10 (3)

Solving (1) and (3), we get x² = 9 and y² = 1

x = ±3 and y = ±1

From (2), x and y are of opposite sign,

\ (x = 3 and y = -1) or (x = -3 or y = 1)

\ z = ± (3 – i)

**2. Calculate: z = √(10-2i)**

Answer:

z1 = √(10-2i) = 3.1778954-0.3146737i = 3.1934369 × ei (-0.0314165)

Complex number: 10-2i

Square root: sqrt (the result of step No. 1) = sqrt(10-2i) = √ 10-2i = 3.1778954-0.3146737i.

Related post: Complex Number | Addition, Subtraction, Multiplication, Division and Exercises

Sources: PinterPandai, Wikipedia