Zalgorithm

Polynomial functions

A polynomial is a mathematical expression consisting of indeterminates (variables) and coefficients. Polynomial expressions can only use the operations of addition, subtraction, multiplication, and exponentiation (to non-negative powers).

“An example of a polynomial of a single indeterminate x”: x24x+7x^2 - 4x + 7.

Polynomial expressions are used to form polynomial equations: f(x)=x24x+7f(x) = x^2 - 4x + 7.

Algebraic equation

The term algebraic equation is used for the univariate case of polynomial equations. A univariate object is an expression, equation, function or polynomial that only has one variable (one indeterminate).1

Solving polynomial equations

To solve a polynomial means to set its equation to some value, usually zero, then find the values that make the expression true. For my current interests, this means finding the “roots” or zeros of the equation — finding the variable values where the expression equals zero.

Solving polynomial equations for complex roots

f(x)=x8+1 f(x) = x^8 + 1 x8+1=0 x^8 + 1 = 0 x8=1 x^8 = -1

Express -1 in polar form

TODO: confirm that this is actually the polar form and not the exponential form. (It’s not technically accurate. See notes / Introduction to complex numbers.)

What’s written below is technically describing the exponential form. Any complex number can be written as reiθr\cdot e^{i\theta}, where:

For -1:

So 1-1 can be written as a complex number as 1eiπ1e^{i\pi}

But 1-1 can also be written as:

1=e(i(π+2πk)) -1 = e^{(i(\pi + 2\pi k))}

This can be done for any integer kk, since adding 2π2\pi doesn’t change the angle.

Expressing 1-1 in the polar form means that x8=1x^8 = -1 can be written as:

x8=e(i(π+2πk)) x^8 = e^{(i(\pi + 2\pi k))}

Take the 8th root

Now solve for xx:

x=ei(π+2πk)/8=eiπ(1+2k)/8 x = e^{i(\pi + 2\pi k)/8} = e^{i\pi(1 + 2k)/8}

Find all distinct roots

For k=0,1,2,,7k = 0, 1, 2,\dots, 7 there are 8 distinct values:

Convert to rectalinear form

The general process

Given a complex number in the exponential form z=re(iθ)z = re^{(i\theta)}, convert it to the rectalinear form z=a+biz = a + bi.

Apply Euler’s formula:

Euler’s formula states that e(iθ)=cosθ+isinθe^{(i\theta)} = \cos\theta + i \sin\theta.

Therefore, z=re(iθ)=r(cosθ+isinθ)z = re^{(i\theta)} = r(\cos\theta + i \sin\theta)

Distribute rr:

z=rcosθ+i(rsinθ) z = r\cos\theta + i(r \sin\theta)

Identify the real and imaginary parts:

Therefore: z=rcosθ+i(rsinθ)z = r\cos\theta + i(r\sin\theta)

Concrete example

Convert z=2e(iπ/4)z = 2e^{(i\pi/4)} to rectalinear form:

Calculate:

To get z=2+i21.414+1.414iz = \sqrt{2} + i\sqrt{2} \approx 1.414 + 1.414 i

See notes / Solve cosine of pi over four for details about what’s going on with the trigonometry.

General form for finding roots of complex numbers

Using f(x)=x30.1f(x) = x^3 - 0.1 as an example.

The general form for finding the roots of complex numbers:

x=r(1/n)e(i(θ+2πk)/n),for k = 0, 1, 2,...,n-1 x = r^{(1/n)}\cdot e^{(i(\theta+2\pi k)/n)}, \text{for k = 0, 1, 2,...,n-1}

For x3=0.1x^3 = 0.1:

x30.1=0 x^3 - 0.1 = 0 x3=0.1=0.1e(i(0+2πk)) x^3 = 0.1 = 0.1e^{(i(0+2\pi k))}

The three roots (n=3) are:

k=0:x0=0.1(1/3)e(i0) k=0: x_0 = 0.1^{(1/3)}\cdot e^{(i\cdot0)}

Remember that e^0 = 1, so x_0 is a real number.

k=1:x1=0.1(1/3)e(i2π/3) k=1: x_1 = 0.1^{(1/3)}\cdot e^{(i\cdot2\pi/3)} k=2:x2=0.1(1/3)e(i4π/3) k=2: x_2 = 0.1^{(1/3)}\cdot e^{(i\cdot4\pi/3)}

References

Wikipedia contributors. “Polynomial.” Wikipedia, The Free Encyclopedia. https://en.wikipedia.org/w/index.php?title=Polynomial&oldid=1324301130.(accessed December 9, 2025).

Notes


  1. Wikipedia contributors, “Univariate,” Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Univariate&oldid=1223573558 (accessed December 10, 2025). ↩︎

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