# Why is there more than one answer for root of complex numbers [closed]

I learnt to find the root of a complex number we should use the equation below: $$\sqrt[n]{Z}=\sqrt[n]{R}.e^{i\frac{\theta+2k\pi}{n}}$$ And $k$ will be $0$ to $n-1$.

So for example a cube root of a complex number will have 3 answers. With $k=0$ being the principal root but why?!

What does it mean to have 3 answers?

Let's find the cube root of $3+4i$, the results are: \begin{align*} k=0 &\implies 1.6289 + 0.5202 i \\ k=1 &\implies -1.2650 + 1.1506 i \\ k=2 &\implies -0.3640 - 1.6708 i \end{align*} These are completely different complex numbers.

## closed as off-topic by Namaste, Taroccoesbrocco, Delta-u, José Carlos Santos, Parcly TaxelJul 16 '18 at 4:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Delta-u, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

• Because all those solutions satisfy the equation $z^3=3+4i$. – Javi Jun 22 '18 at 21:42
• Real numbers cave have multiple roots, too. For example, $4$ has both $-2$ and $2$ as square roots. Don't confuse an $n$-root of a number with the n-th root function. – Xander Henderson Jun 22 '18 at 21:45

Well, first you should realize that what is meant by the $n$-th root of a complex number $z_0$ is a complex number $z$ solving the equation $$z^n=z_0\;.$$ It seems that you understand well that there are $n$-solutions to this equation.

I infer that your confusion arises in defining the $n$-th root function $z\mapsto \sqrt[n]{z}$. The problem here being that this is not a well defined function on the complex numbers. Actually, this ambiguity arises in the real numbers as well. For example, for any real number $x$, there are two square roots $\sqrt x$ and $-\sqrt{x}$, both solve the equation $y^2=x$.

Just because there is ambiguity doesn't mean we can't define a function, however. For example, in the real case, we can take the positive square root and define the function $x\mapsto \sqrt{x}$. We pay a price though for this choice though -- it now has a restricted domain (restricted to the positive reals).

In the complex setting the situation is similar. For example, take the point $z=1$, you can extend the function $z\mapsto \sqrt[n]{z}$ to a single valued function in a small neighborhood around $z=1$. However, you can not extend this function globally (on all of $\mathbb{C}$) to a single, well defined, $n$-th root function.

In complex analysis, such functions are called multi-valued. They are local sections of branched coverings.

Let's start with a cubic equation whose roots are all real. If you picture graphing a function of the form $$y = a x^3 + b x^2 + cx + d$$ with $a$ positive, you can immediately say that $y$ will be negative for very large negative $x$ (where the $x^3$ term is much bigger than $x^2$) and positive for very large positive $x$, so the curve crosses zero at least once. Then if you picture letting $c$ be some large enough negative number, you can deduce that the slope will be negative near $x=0$. So you will have a maximum and a then a minimum, before proceeding toward $x = +\infty$.

Finally, depending on the value of $d$, the maximum and minimum might be on opposite sides of $y=0$. In that case, there are three real solutions to the equation.

Now picture changing $a,b,c,d$ slowly, so that one of the max or min goes to the other side of the $x$ axis. Two of the real solutions go away. But just like for a quadratic equation, they will be replaced by a pair of complex solutions. The point is, you still have three solutions, or more accurately, the cubic has three factors of the form $(x-s_i)$ where each $s_i$ is a real or complex number.

The cubic equation $x^3 = 1$ is not an exception to this rule. So there must be three numbers such that $$x^3-1 = (x-s_1)(x-s_2)(x-s_3)$$ Perhaps two of those $s_i$ could be equal, but there are still three linear factors.

Those three cube roots of $1$ you encounter are just the $s_i$ in those three those factors.

(I would have called the variables $r_i$ but did not want to "give away" the point of the argument by naming them as the "roots" arbitrarily!)

In fact, every polynomial of degree $d$ has a factorization into $d$ linear terms $(x-s_i)$ as long as you allow the $s_i$ to be complex, and this holds even if the coefficients in the polynomial are themselves complex. (Thus there is no need to go beyond the ordinary complex numbers to add in $\sqrt{i}$.)

That statement just made is called the "Fundamental Theorem of Algebra" and is somewhat non-trivial to prove. In fact Gauss, arguably the greatest mathematician ever, is credited by many with producing the first good proof of this theorem.