When do you take into account the +2kpi for complex numbers arguments in complex equations

Question

$$z^2 = ({2e^{i{\frac{\pi}{3}}}})^8$$

To find z I took the square root of both sides which gives me:

$$z = ({2e^{i{\frac{\pi}{3}}}})^4$$ which I rewrote as $$z = {2^4e^{i{\frac{4\pi}{3}}+2k\pi}}$$

and for k=0 we find $$z=-8-8\sqrt(3)$$ which is one of the correct answers but for k=1 we find the same thing. On the other hand the correction states the other answer is $$z=8+8\sqrt(3)$$

I think that I need to put the $$+2k\pi$$ before taking the square root of Z but I don't understand why it changes the answer since mathematically it should be the same?

Because even when I tried with $$z = ({2e^{i{\frac{\pi}{3}}+2k\pi}})^4$$ which I rewrote as $$z = {2^4e^{i{\frac{4\pi}{3}}+8k\pi}}$$ I still don't find the correct second answer.

Thanks

Answer 1

Okay. Let's beat this horse to death and then beat it some more.

We know: $e^{2k\pi i} = 1$ and $e^{(2k + 1)\pi i} = -1$ and therefore we know $e^{M + 2k\pi i} = e^{M}e^{2k\pi i} = e^{M}$ and $e^{M + (2k + 1)\pi i} =e^{M}e^{(2k+1)\pi i} = -e^{M}$.

So when we write something like $z = re^{\theta i}$ we also know that $z = re^{\theta i + 2k\pi i}$. But we don't have to write that as it is understood.

When we rise complex numbers to integer powers we have $z^n = (re^{\theta i})^n = (re^{\theta i + 2k\pi i})^n$ or $z^n = r^ne^{n\theta i} = r^ne^{n\theta i + 2(kn)\pi i}$. But as $2(kn)$ is also an even integer we don't have to write that as it is understood.

(This shouldn't surprise us. If $x = a$ then $x^n$ will only have one value.)

But when we raise things to a fractional power things do matter and we get multiple results. (Which shouldn't surprise us. If $x = a^n$ then $a = $ one of the $n$ $n$th roots of $x$.) If $z = re^{\theta i} = re^{\theta i + 2k\pi i}$ then $z^{1/n} = r^{\frac 1n} e^{\frac{\theta}ni + \frac{2k}n \pi i} = [r^{\frac 1n} e^{\frac{\theta}ni}]*e^{\frac{2k}n \pi i}$.

Now $0< \frac {2}n\pi < \frac {4}n\pi < .... < \frac {2(n-1)}n\pi$ or distinct non equal values for $k = 0..... n-1$ but for $k \ge n$ we have $\frac {2k}n \pi = \frac {2(k-n)}n\pi + 2\pi$ is not distinct, we have precisely $n$ distinct values of $n$th roots of $z$.

It's worth noting if $n= 2$ we have $z^{1/2} = \sqrt{r}e^{\frac{\theta} 2i + \pi i} = \{\sqrt{r}e^{\frac{\theta} 2i}, \sqrt{r}e^{\frac{\theta} 2i+ \pi i}\} = \{\sqrt{r}e^{\theta i/2 }, - \sqrt{r}e^{\theta i/2 }\} = \pm \sqrt{r}e^{\theta i/2}= \pm \sqrt{z} $. Which is our old familiar rule: $a^2 =z \implies a = \pm \sqrt{z}$.

In fact we can extend that: if $a^n = z = re^{\theta i}$ and we define $\sqrt[n]{z} = \sqrt[n]{r}e^{\theta i/n}$ then $a = \sqrt[n]{z}*\omega$ where $\omega = e^{\frac{2k}n \pi i}; k=0....(n-1)$. And we can further note that the $n$ distinct $\omega$s are the $n$ distinct $n$-th roots of $1$.

So now we've kill the horse. Let's keep beating it....

Take your problem:

$z^2 = ({2e^{i{\frac{\pi}{3}}}})^8$

we can do what you attempted and it will work:

$z = \pm({2e^{i{\frac{\pi}{3}}}})^4=$

$z = \pm 16e^{i\frac{4\pi}{3}}= \pm 8 \pm 8\sqrt{3}i $

or we could do it the thourough way:

$z^2 = (2e^{i\frac{\pi}{3}+ i 2k\pi})^8 = 2^8e^{i\frac{8\pi}3 + i16k\pi}$

$z = \pm2^4e^{i\frac{4\pi}3 + i8\pi} = \pm16e^{i\frac{4\pi}3}=\pm 8 \pm 8\sqrt{3}i$.

Answer 2

1. You can use this method:

$$z^2=(2e^{i{\pi\over3}})^8=(2^8e^{i{8\pi\over3}})=(2^8e^{i{8\pi\over3}+2ki\pi})$$

and now you take the square root.

2. Otherwise you can take the square root putting $\pm$:

$$z=\pm(2e^{i{\pi\over3}})^4$$

3. We can also write $2e^{i{\pi\over3}}=2e^{i{\pi\over3}+2li\pi}$ because any complex number is defined up to a phase factor $e^{2li\pi}$, so:

$$z^2=(2e^{i{\pi\over3}+2li\pi})^8=(2^8e^{i{8\pi\over3}+16li\pi})=(2^8e^{i{8\pi\over3}+16li\pi+2ki\pi})=(2^8e^{i{8\pi\over3}+2\pi i(8l+k)})$$ $$z=2^4e^{i{4\pi\over3}+\pi i(8l+k)}=2^4e^{i{4\pi\over3}}e^{\pi i(8l+k)}=\pm2^4e^{i{4\pi\over3}}$$

with $l,k\in\mathbb Z$, but $8l+k$ can be even or odd so you get the same solutions.

Answer 3

We already know, from real number arithmetic, that taking roots of real numbers is not well-defined: every positive real number has two square roots, two fourth roots, and in general two $k$-th roots for every even integer $k \ge 2$.

The situation in the complex numbers is even worse (or better, depending on your perspective). Every nonzero complex number has two square roots, three cube roots, four fourth roots, etc. Those roots are found using the "$+2k\pi$" method referred to in your question.

You can think of the method like this: when solving an equation like $z^2 = W$, first multiply the right hand side by $e^{2k\pi i}$ to get $z^2 = W e^{2 k \pi i}$; then you may manipulate the exponents as you are used to doing.

So, before you even wrote "I took the square root of both sides", the $+2k\pi$ should have kicked in. Instead you should have written: $$z^2 = (2e^{i\frac{\pi}{3}})^8 \cdot e^{2 k \pi i} $$ and now you are ready to "take the square root" by manipulating exponents. The alternate answer will come from setting $k=1$.

Added to address the comment: The reason this works is because we can explicitly list the two square roots of $1$ using the $+2k\pi$ method. They are: $$e^{(2\pi i)/2} = -1, \quad e^{(4 \pi i)/2} = 1 $$ So, if you've already found one square root of $z$, the other one can be found by multiplying by $-1$.

At the risk of confusing the issue, but with the hope of clarifying it, perhaps you might consider the similar situation with cube roots. The three cube roots of $1$ are $$e^{(2 \pi i)/3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2} $$ $$e^{(4 \pi i)/3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2} $$ $$e^{(6 \pi i)/3} = 1 $$ So, if you have already found one cube root of a complex number, the other two cube roots can be found by multiplying by $-\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$.