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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.

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

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    $\begingroup$ Because all those solutions satisfy the equation $z^3=3+4i$. $\endgroup$ – Javi Jun 22 '18 at 21:42
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    $\begingroup$ 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. $\endgroup$ – Xander Henderson Jun 22 '18 at 21:45
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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.

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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.

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