Complex Number Confusion (1 Viewer)

X-burner · Nov 5, 2011

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 · Nov 5, 2011

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 · Nov 5, 2011

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 · Nov 5, 2011

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

Let us consider,

$z_{1}=r_{1}(\cos\theta_{1} + i\sin \theta_{1})$

$\arg z_{1}=\theta_{1}$

$z_{2}=r_{2}(\cos\theta_{2} + i\sin \theta_{2})$

$\arg z_{2}=\theta_{2}$

What if we were to multiply these two together?

$z_{1}z_{2}=r_{1}r_{2}(\cos\theta_{1} + i\sin \theta_{1})(\cos\theta_{2} + i\sin \theta_{2})$

$=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+\cos\theta_{1}i\sin \theta_{2}+ i\sin \theta_{1}\cos\theta_{2}-\sin \theta_{1}\sin \theta_{2})$

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

$=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}-\sin \theta_{1}\sin \theta_{2}+i(\sin \theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin \theta_{2}))$

$=r_{1}r_{2}(\cos(\theta_{1}+\theta_{2})+i(\sin \theta_{1}+\sin \theta_{2}))$

What happened to the arguments now?

$\arg z_{1}z_{2}=\theta_{1}+\theta_{2}=\arg z_{1} + \arg z_{2}$

We can see that the arguments have added!

That explanation was for the general case, so if-

$z=r(\cos \theta + i\sin \theta)$

$\arg z = \theta$

$z^2=r^2(\cos (\theta+\theta) + i\sin (\theta+\theta))$ (Arguments added as we have shown before.)

$=r^2(\cos 2\theta+i\sin 2\theta)$

$\therefore \arg z^2 = 2\theta =2\arg z$

Have a look at this trend, once again,

$z=r(\cos \theta + i\sin \theta)$

$\arg z = \theta$

$z^3=r^3(\cos (\theta+\theta+\theta) + i\sin (\theta+\theta+\theta))$

$=r^3(\cos (3\theta) + i\sin (3\theta))$

$\therefore \arg z^3 = 3\theta=3\arg z$

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

$\arg z^n = n\arg z$

math man · Nov 5, 2011

or by de moivre's theorem:

let $z = cosx + isinx$

taking both sides to the power of n:

$z^{n} = cos(nx) + isin(nx)$

now $arg(z^{n}) = nx$

$arg(z)= x$

hence we get:

$n[arg(z)]= nx$

Therefore: $n[arg(z)]= arg(z^{n})$

Complex Number Confusion (1 Viewer)

X-burner

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bleakarcher

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bleakarcher

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SpiralFlex

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math man

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