Convert sqrt(27) + 3i to Re^{iθ}: Guide

[Lancelot59]

The problem is that I need to convert:

$$\sqrt{27} + 3i$$

From the form $$(a+bi)$$ to $$Re^{i\theta}$$. I have no clue what to do with this. I do know the following:

$$e^{i\pi}=cos(\theta)+sin(\theta)i=-1$$

But I don't see how that's helpful. This is the first of several problems.

 

[SVXX]

What you need to know is that for a complex number z = a + ib, we define the following -:

and

Now can you complete this question?

Sidenote : r is called the "modulus" or "magnitude" of the complex number and theta is its "amplitude" or the "argument".

 

[Lancelot59]

What you need to know is that for a complex number z = a + ib, we define the following -:

and

Now can you complete this question?

Sidenote : r is called the "modulus" or "magnitude" of the complex number and theta is its "amplitude" or the "argument".

I see here. Correct me if I'm wrong here:

An imaginary number is like a vector, with a real and imaginary component. Am I correct in saying that the new vector:
$$6e^{i\frac{\pi}{6}}$$
Is now in "polar coordinates"? Or some form of polar coordinates with an imaginary axis?

Another question, is modulus just a term used for the magnitude, or is there more to it than that? I've always known it to be the remainder of division.

 

[SVXX]

Indeed, it is in polar coordinates in the exponential form.

And

which is why it is called the "modulus" of a complex number (its how the modulus function is defined for complex numbers).

 

[Lancelot59]

Indeed, it is in polar coordinates in the exponential form.

And

which is why it is called the "modulus" of a complex number (its how the modulus function is defined for complex numbers).

Okay, now I have one where I need to go the other direction:

$$2e^{\frac{2\pi}{3}i}$$

So I got the following:

$$tan(\frac{2\pi}{3})=\frac{b}{a}=-\sqrt{3}$$
$$b=-a\sqrt{3}$$

$$\sqrt{a^{2}+b^{2}}=2$$
$$a^{2}+b^{2}=4$$
$$a^{2}+(-a\sqrt{3})^{2}=4$$
$$a^{2}+3a^{2}=4$$
$$4a^{2}=4$$
$$a^{2}=1$$
$$a=+-1$$

I'm not sure where to go from here.

EDIT: Nevermind, I was looking at the wrong answer in the answer key.