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How to find $|z|$ and $\arg (z)$

$z$ is complex number and $z$ is defined by $$z=\left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\cdot(3-3i)^{20}$$

I`ve tried to behave it like $$e^{i15\frac\pi5}\cdot e^{i20\frac\pi4}$$ and got in result = $$e^{i3\pi}\cdot e^{i5\pi}=e^{i8\pi}$$ which gives me $$(-1)^8=1$$ and if $$+z=|z|$$ so $$|z|=1$$ So is it my answer and how to find? $$Arg(z)$$

Rodrigo de Azevedo
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Ulan Duishenaliev
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 25 '22 at 18:04
  • Hint: what do you know about arg and absolute value of a product. – Robert Israel Apr 25 '22 at 18:11
  • @Community I think it's clear that they're asking us to help the find the mod and arg? – Pineapple Fish Apr 25 '22 at 18:30
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    But $,3-3i \ne e^{i,\pi/4},$. – dxiv Apr 25 '22 at 18:31
  • @dxiv I solved it by this way: the angle between (x,y)=(3,-3) is pi/4. Then 3(cos pi/4 + sin pi/4)^20=e^20pi/4 – Ulan Duishenaliev Apr 25 '22 at 18:36
  • Yeah think of it on an argand diagram. You're on the right track but $3-3i$ lies $45^\circ$ from the x-axis, which means that for it, $\theta=\frac{-\pi}{4}$. But this only gives you something with modulus 1, whereas you need something with the same modulus as $3-3i$, which by the pythagorean theorem, has a modulus of $3\sqrt{2}$. Multiplying by this real number to scale it up by that exact quantity, we get that $3-3i$ actually equals $3\sqrt{2}e^{-i\frac{\pi}{4}}$, if I express it in symbols. Remember that the moduli multiply and the arguments add. I wish I could draw a picture to help – Pineapple Fish Apr 25 '22 at 18:36
  • @UlanDuishenaliev $;3-3i=3\sqrt{2}\big(1/\sqrt{2} - i / \sqrt{2}\big)=3\sqrt{2}\big(\cos(-\pi/4) + i \sin(-\pi/4)\big),$. – dxiv Apr 25 '22 at 18:39
  • @UlanDuishenaliev ahh, $3-3i$ is in the fourth quadrant. You have to rotate negative $\theta$'s to get to it – Pineapple Fish Apr 25 '22 at 18:44
  • @dxiv so then it will be equal to? $$e^{-i5pi}$$ – Ulan Duishenaliev Apr 25 '22 at 18:49
  • @PineappleFish so if I understood correctly I`ll get in answer e^{i3pi} * -3*square root of(-2(-1)^{3\4)) – Ulan Duishenaliev Apr 25 '22 at 19:10
  • @UlanDuishenaliev You lost a factor of $3 \sqrt{2}$ along the way. – dxiv Apr 25 '22 at 21:36

2 Answers2

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This is just a starter to actually answer the question, I'd be willing to have a longer discussion in comments or chatroom;

$\left(\cos(\frac{\pi}{5})+i\sin(\frac{\pi}{5})\right)^{15}(3-3i)^{20}=(e^{i\frac\pi5})^{15}\left(3\sqrt{2}e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}(3\sqrt{2})^{20}\left(e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{20\cdot-i\frac\pi4}\right)=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{-5\pi i}\right)=3^{20}\cdot\sqrt{2}^{20}e^{-2\pi i}$

but note that $3^{20}\cdot\sqrt{2}^{20}$, but again by laws of exponents you can simplify further because $\sqrt{2}=(2)^{\frac12}$, so $3^{20}\cdot\sqrt{2}^{20}=3^{20}\cdot\left((2)^{\frac12}\right)^{20}=3^{20}\cdot2^{\left(20\cdot\frac12\right)}=3^{20}\cdot2^{10}$

so the final answer is $3^{20}\cdot2^{10}$, or 3570467226624. This is a pure real number with imaginary part 0. It's so ludicrously large, but this actually makes sense conceptually because the modulus of $3-3i$ (which for goodness sake you're taking a 20th power of) is $\sqrt{3^2+3^2}=3\sqrt{2}\approx4.2426$, and so with each multiplication you increase the modulus by a constant factor, essentially blowing it up exponentially.

Also this is a cool video: https://www.youtube.com/watch?v=-dhHrg-KbJ0&t=531s

You'll notice that part of the entire point of Euler's Formula is that it's one possible parametrization of the unit circle. There is further intuition on Euler's Formula in this video too, although its a bit long winded and I haven't watched it in a while.

To me one of the craziest parts about Euler's formula is the seemingly ridiculus connection between rotation, and taking exponents. Like literally, that's just straight up crazy.

Also, this might be of interest to you:

$\left(re^{i\theta}\right)^n=r^ne^{in\theta}$

From this, we can draw some general conclusions:

  • Just as is the case with real numbers, if $r<1$, then $r^n$ (for high powers of $n$) become very tiny. If $r=1$ then everything about $r^n$ stays fixed, and you have a pure rotation. Finally if $r$ where greater than 1 - even if $r$ is only slightly bigger like 1.000000001 with lots of zero's, with a high enough power of $n$ you'd expect to get complex numbers with a very large modulus.

  • $e^{it}$ is periodic, so the result can point in any direction depending purely on the value of $t$

Interestingly, Euler's Theorem is also super handy for taking square roots of complex numbers too, since square roots are basically just another name for raising to the one-halfth power, after all;

$\left(re^{i\theta}\right)^\frac12=r^\frac12e^{i(\theta/2)}$

which also pretty ridiculous. I've made you a Desmos thing to play around with: https://www.desmos.com/calculator/dvfoycjgz8 . The red circle to pick a $z$, and the slider $s$ is the exponent. As you can see you can always pick values of the power that loop you all the way back to numbers where the real part is the only part. It just so happens to be that in a lot of the questions you see in school they give you problems that are more likely to work out nicely, although this is not generally the case.

Pineapple Fish
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Just remember some basic properies of $\arg$ and absolue value: If a value is given in polar form, then the polar coordinates are obtained. With $r, \varphi\in \Bbb R$ and $r\geqslant 0$:

$$\begin{align*} |r\cdot e^{i\varphi}| &= r \\ \arg (r\cdot e^{i\varphi}) &\equiv \varphi \end{align*}$$ Where $\equiv$ means that the value is only determined up to an integer multiple of $2\pi$. Common choices for $\arg$ are $0\leqslant\arg<2\pi$ or $-\pi < \arg\leqslant \pi$.

Moreover, for a product of two complex numbers we have:

$$\begin{align} |z\cdot w| &= |z|\cdot|w| \\ \arg(z\cdot w) &\equiv \arg z + \arg w \\ \end{align}$$ which implies for $z\neq 0$ and real exponents $p\neq 0$ $$\begin{align} |z^p| &= |z|^p \\ \arg(z^p) &\equiv p\cdot\arg z\\ \end{align}$$

So the first lines for your $\arg z$ would prepare for easy computation and read:

$$\begin{align} z &= \left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\!\!\cdot\,(3-3i)^{20} \\ &= \left(\exp\frac{\pi i}5\right)^{15}\!\!\cdot\,3^{20}\cdot(1-i)^{20} \\ \end{align}$$ Hence: $$\begin{align} \arg z &\equiv 15\cdot \frac\pi5 + 20\cdot\underbrace{\arg(3)}_{\textstyle=0} + 20\cdot\underbrace{\arg(1-i)}_{\textstyle=-\pi/4} \\ &\equiv 3\pi - 5\pi \equiv 0 \mod 2\pi \end{align}$$

emacs drives me nuts
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