Complex Number Confusion (1 Viewer)

[X-burner]

I don't get how arg(z^n) = n arg (z), do u just remember it, can someone prove by induction or just explain what happened there. Thanks

 

[bleakarcher]

Define the proposition P(n): arg(z^n)=n*arg(z), n integral
When n=1, LHS=arg(z)=RHS
Hence, P(1) is true
Assuming P(k) is true,
ie P(k):arg(z^k)=k*arg(z), k integral
Try to prove for P(k+1):
P(k+1):arg(z^[k+1])=(k+1)arg(z)
Now, LHS=arg((z^k)*z)=arg(z^k)+arg(z)=k*arg(z)+arg(z)=(k+1)arg(z)=RHS
Hence, if P(k) holds P(k+1) holds also
Since P(1) is true, P(2) must be true and so on for all integers n>0

 

[bleakarcher]

btw dude if u dont understand my prove for LHS=RHS, here r the laws i assumed u already knew:
z^(k+1)=(z^k)*(z^1) as (x^m)*(x^n)=x^(m+n)
also:
arg[z1*z2]=arg(z1)+arg(z2)

 

[SpiralFlex]

Another explanation (To help you understand what is going on.)

Let us consider,













What if we were to multiply these two together?







We get a semi-messy expansion, so if I arrange it to a familiar form of,







What happened to the arguments now?



We can see that the arguments have added!



That explanation was for the general case, so if-






(Arguments added as we have shown before.)







Have a look at this trend, once again,
















And if we to do this so on, the general case is,

 

[math man]

or by de moivre's theorem:

let

taking both sides to the power of n:



now



hence we get:



Therefore: