2e^5i in rectangular form?

I hear of the e^ia = (-1)^a/pi
But I'm not sure what -1^5/pi has in relation to 2e^5 like, is there a relation on polar coordinates that I'm missing to I somehow get this in the form:
x + yi by rearranging?

Reply 1

simon0

It can be obtained from that:

e^{ix} = \cos (x) + i \sin (x),

R e^{ix} = R \big( \cos (x) + i sin (x) \big),

where for a complex value:

z = a + ib,

where a and b are real values:

\bullet \, \, R = \lvert z \rvert = \sqrt{a^{2} + b^{2}},

(can be seen as the norm of z),

\bullet \, \,

x is the argument of z (the angle z makes with the positive real axis).

--------------------------------------------------------------------------------------

So in effect a complex value z has three forms (not including z=0).

Does that clear up a few things?

(edited 3 years ago)

Reply 2

NeedABtterUserID

Original post by simon0

It can be obtained from that:

e^{ix} = \cos (x) + i \sin (x),

R e^{ix} = R \big( \cos (x) + i sin (x) \big),

where for a complex value:

z = a + ib,

where a and b are real values:

\bullet \, \, R = \lvert z \rvert = \sqrt{a^{2} + b^{2}},

(can be seen as the norm of z),

\bullet \, \,

x is the argument of z (the angle z makes with the positive real axis).

--------------------------------------------------------------------------------------

Does that clear up a few things?

so are is the coefficient (in my case 2) but how does the ix collapse into separate cos and sin?

Reply 3

simon0

Few examples include:

- z = 1 + i:

R = \sqrt{1^{2} + 1^{2}} = \sqrt{2}, \, \, x = \mathrm{arctan} (1) = \frac{ \pi }{4}

,

- z = 1 + 2i:

R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}, \, \, x = \mathrm{arctan} (2) = 1.1071 \ldots

,

- z = -2 + 4i:

R = \sqrt{2^{2} + 4^{2}} = 2 \sqrt{5}, \, \, x = \frac{ \pi }{2} + \mathrm{arctan} \Big( \frac{2}{4} \Big) = 2.03 \ldots

Reply 4

NeedABtterUserID

OK but that, forgive me if I'm wrong ends in polar form and somes from a simple complex number of a + ib.
How the hell do I apply that to this question?????
Cartesian form is required, I just dont know how to get it from an exponential as it is a whole equation unto itself.

(edited 3 years ago)

Reply 5

simon0

Original post by NeedABtterUserID

so are is the coefficient (in my case 2) but how does the ix collapse into separate cos and sin?

One way is to use the Taylor expansion of the exponential with "ix" as the argument.

Then you can separate the terms to obtain the Taylor expansion for sin(x) and icos(x).
This is well done from website:

https://socratic.org/questions/how-do-you-show-that-e-ix-cosx-isinx

Reply 6

NeedABtterUserID

Original post by simon0

OK, so I do this in order to get it into polar form in order to convert into Cartesian form?

Reply 7

simon0

Original post by NeedABtterUserID

I do not understand the first two lines of your reply.

Of your last line, it looks like you want to go from the exponential form to the Cartesian form.
Just use the formula I gave earlier:

z = R e^{ix} = R \cos (x) + i R \sin(x) .

For example:

z = \sqrt{2} e^{i ( \pi / 4)} = \sqrt{2} \cos( \pi /4) + i \sqrt{2} \sin ( \pi /4) = 1 + i .

Is that okay? :smile:

(edited 3 years ago)

Reply 8

NeedABtterUserID

Original post by simon0

I do not understand the first two lines of your reply.

Of your last line, it looks like you want to go from the exponential form to the Cartesian form.
Just use the formula I gave earlier:

z = R e^{ix} = R \cos (x) + i R \sin(x) .

For example:

z = \sqrt{2} e^{i ( \pi / 4)} = \sqrt{2} \cos( \pi /4) + i \sqrt{2} \sin ( \pi /4) = 1 + i .

Is that okay? :smile:

I think I understand now, thanks a million!
May I just verify I have this right:
For my case I have Re^ix where x is a whole number not a multiple or division of pi, does this mean that I would plug a whole number into my sin and cos?
This just seems strange because the number is quite high its 19. (the real equation is 25/27e^19i (I wanted to understand not just steal answers)) but this would suggest multiple rotations around the origin, do I just divide by pi and take the remainder?

Reply 9

simon0

Original post by NeedABtterUserID

You are right, the value x is substituted into the sin and cos functions in the formula.

19 seems high but is valid but substitute as usual (then afterwards we can find the argument/x value which is between

- \pi

\pi

which this x-value is better known as the "Principal Argument" but we are getting ahead here).

You are correct it does mean multiple rotations around the origin.

Fun fact:
As all complex values (except 0) can have various representations for example: