How do you get 2Theta from z^2 ? (1 Viewer)

[Starcraftmazter]

In the question,
Show that arg (1-i)z^2 = 2

- Pie/4+-2Pie

Now, (1-i) = Pie/4 is very simple, and the +-2pie thing is all straightforward, using the arg(z1.z2) = arg (z1) + arg (z2)

But how the heck do you get 2 Theta from z^2 ???


Thanks

 

[Starcraftmazter]

On line 3, isn't that r^2Sin2@?

And what happens to the Sin, to only have 2@, without the sin?

 

[Wackedupwacko]

arg( r^2cis2@) means the argument of r^2 cis2@ which is just 2@ (the angle only)

this is one of the basics of complex numbers. argument is only the angle whilst modular is the r^2 bit...

 

[Starcraftmazter]

Yeh, but you're not given 2@ to start off with, you're gizen z^2, which is r^2(cis@)^2, not r^2.cis2@

So I just went ahead and proved it this way:

z^2 = r^2 . (Cos @ + i.Sin @)^2
= r^2 . ( Cos^2 @ + 2.Sin@.Cos@.i + Sin^2 @.i^2 )
= r^2 . ( Cos^2 @ - Sin^2 @ + i . Sin 2@)
= r^2 . ( Cos 2@ + i . Sin 2@)

Now arg (z^2) = tan^-1 (sin 2@ / Cos 2@)
Therefor, arg (z^2) = tan^-1 (tan 2@)
= 2@


Which works, so yeh 8)

 

[pLuvia]

z²
=r²(cis@)²
=r²cis2@, using DeMoivre's theorem

 

[Starcraftmazter]

Yeh, I don't think we've learned that quite yet.

 

[Wackedupwacko]

if u havent learnt it yet then just take our word for it.

demoives is

cis^n @ = cisn@

its proven by mathematical induction if u want the proof ill write it down later

 

[Affinity]

expand

(cos[t] + i sin[t])(cos + i sin)

you should get

cos[t+s] + i sin[t+s]

by using suitable trig identities

then it's obvious

 

[buchanan]

Even more obvious if you use Euler's formula e^iθ=cis(θ )

cis(t+s)=e^i(t+s)=e^ite^is=cis(t).cis(s).

As for De Moivre's Theorem,

(cis(θ ))ⁿ=(e^iθ)ⁿ=e^inθ=cis(nθ ) and true for all real n, not just integers n (better than induction).