Complex Numbers finding values

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 3K views
lunds002
Messages
21
Reaction score
0

Homework Statement



Given that arg((z1z2)/2i) = [itex]\pi[/itex], find a value of k.

Homework Equations



arg z2=2[itex]\pi[/itex]=0

arg z1=[itex]\pi[/itex]/6


The Attempt at a Solution



(([itex]\pi[/itex]/6)^k x (2[itex]\pi[/itex]))/ (2i) = [itex]\pi[/itex]

I'm not sure what to do with the imaginary number i..
 
Physics news on Phys.org
lunds002 said:
Given that arg((z1z2)/2i) = [itex]\pi[/itex], find a value of k.

You're missing a k in the problem. Where is it missing?
 
Oops, should be arg ( z1 x z2^k) / (2i)
 
lunds002 said:
Oops, should be arg ( z1 x z2^k) / (2i)

You should be able to get it by expanding the arg function.

E.g. arg(z1z1)=arg(z1)+arg(z2).

(the rules are similar to log rules)
 
Okay so then I get

arg(z1^k) + arg(z2) = pi
2i

(pi/6)^k + 2pi = pi
2i

(pi/6)^k + 2pi = 2i(pi)

Not sure what to do with the imaginary i
 
lunds002 said:
Okay so then I get

arg(z1^k) + arg(z2) = pi
2i

(pi/6)^k + 2pi = pi
2i

(pi/6)^k + 2pi = 2i(pi)

Not sure what to do with the imaginary i

Check the wikipedia page on argument for how to deal with arg(z1/z2). It's similar to how you expand loga(x/y).

You should know what arg(i) is equal to.
 
Okay so I know z1/z2 = r1/r2 cis (theta-[itex]\psi[/itex])

I don't really understand how that applies here though

And you're right, I do know that arg(i) = -1
 
lunds002 said:
Okay so I know z1/z2 = r1/r2 cis (theta-[itex]\psi[/itex])

I don't really understand how that applies here though

And you're right, I do know that arg(i) = -1

http://en.wikipedia.org/wiki/Argument_(complex_analysis)#Identities

arg(i) is the angle formed by the z=i and the positive real axis. z=i is the line perpendicular to the positive real axis, so arg(i) is not -1 but ?