# Complex Numbers (Principal Argument)

• Nov 17th 2010, 09:08 AM
MattWT
Complex Numbers (Principal Argument)
Hi,

Say you had a function of:

e^((i pi)/7) e^(pi i)

And you had to find Arg(z),

I'm getting a bit lost, i'm trying to convert it into x + iy form, so i can use tan^-1(y/x) to get the final answer,

so im using e^((8 pi i )/7) = cos(8pi/7) + i sin(8pi/7)

but I can't take it down any further, any help would be appreciated, thanks.
• Nov 17th 2010, 09:15 AM
DrSteve
e^((i pi)/7) e^(pi i) = e^(i8pi/7) (just add the exponents).

So arg z = 8pi/7 + 2npi (n an integer)

Arg z is the value of arg z that is between -pi and pi. This is 8pi/7 - 2pi = -6pi/7
• Nov 17th 2010, 10:09 AM
MattWT
So, if I had (2 + 2i) e^(i 5pi/6 )

Then I would take 5pi/6 +2npi

Let n = -1

So 5pi/6 - 2pi = -210deg?

i also expanded it all, getting tan^-1(x/y) = 5pi/12
• Nov 17th 2010, 10:55 AM
DrSteve
No! You first have to write 2 + 2i in exponential form, and then you can do a computation similar to what I did above.

Give it a try, and I'll help you out if you get stuck.
• Nov 18th 2010, 03:41 AM
HallsofIvy
Quote:

Originally Posted by MattWT
So, if I had (2 + 2i) e^(i 5pi/6 )

Then I would take 5pi/6 +2npi

I'm very concerned here. You are asking about exercises using the polar form in the complex plane, yet you seem to have no clue what that means. The point (2, 2), representing 2+ 2i in the complex plane, has modulus (distance from (0, 0)) of $\sqrt{2^2+ 2^2}= \sqrt{8}= 2\sqrt{2}$ and argument (angle with the positive real axis) $\pi/4$: (2+ 2i)e^{i 5\pi/6)= 2\sqrt{2}e^{i \pi/4}e^{i 5\pi/6}= 2\sqrt{2}e^{i(\pi/4+ 5\pi/6}= 2\sqrt{2}e^{i(3 \pi/12+ 10\pi/12}= 2\sqrt{2}e^{13 \pi/12}[/tex].
Since the problem asks for the principal argument, there is no need at all for the " $2n\pi$". The principal argument is with n= 0.

Quote:

Let n = -1

So 5pi/6 - 2pi = -210deg?
Never, ever, convert problems like these to degrees!

Quote:

i also expanded it all, getting tan^-1(x/y) = 5pi/12