Question

"Write Z base360(60), Z base400(150), Z base100(12.5), Z basepi(1/3), in standard form for a complex number (a + ib where a and b are real numbers)"

I know about complex numbers, I'm just having trouble on converting these into complex numbers.

Answer 1

I'm not familiar with the notation... do you have a definition for \(\large Z_{360}(60)\) and the others?

Answer 2

I do, but unfortunately it's kind of poor. This has to do with wrapping functions. The base is interpreted as degrees (360), radians (pi), or gradians (400) on the Unit Circle. Basically, the bases for all of these wrap completely around the Unit Circle. And the part in the parenthesis is a "slice" of the Unit Circle. That's my basic understanding, and the explanation provided for me in my notes.

Answer 3

Okay so based on some quick searching, it looks like you define a function \(Z:\mathbb{R}\to\mathbb{R}^2\) or \(Z:\mathbb{R}\to\mathbb{C}\), i.e. a point on the real line to a coordinate point. Something like that...

So if I'm understanding this correctly, you'd have
\[Z_{360}(60)=1+i\sqrt3\]

Answer 4

I think you're right. Thank you so much!

Answer 5

The others are a bit confusing... I've never worked with gradians before, but even the radians one looks weird to me.

Answer 6

Gradians isn't used very often, and I think the radians one might have been a typo. My professor usually uses 2pi, not just pi. And this one Z base100(12.5) is based on the metric system. It's a new type of math he's trying to bring in.

Answer 7

Hmm okay. I think converting from one unit to another helps visualize what's going on.
\[400\text{ grad}=2\pi\text{ rad}~~\Rightarrow~~150\text{ grad}=\frac{3\pi}{4}\]
and so \[Z_{400}(150)=\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}=-\dfrac{1}{\sqrt2}+i\dfrac{1}{\sqrt2}\]
For the radians one, I think you have
\[Z_{2\pi}\left(\frac{1}{3}\right)=\cos6\pi+i\sin6\pi=1+0i\]
but I could be wrong.

I'm not sure at all about the metric one...

Answer 8

The metric one is more confusing, so I'll probably ask my professor for help on that one.

Answer 9

Actually, I do have an idea: