help please! (complex numbers) (1 Viewer)

[blackfriday]

i can do this question by trial and error but i dont know the proper way...

z=1+isqrt3. find the smallest positive interger for which z^n is real and evaluate z^n for this value of n. show that there is no integral value of n and for which z^n is imaginary.

and...

prove that Iz^nI=IzI^n and arg(z^n)=nargz for all positive intergers n by mathematical induction.

 

[Estel]

z=2cis(pi/3)
for z to be real, z=|z|cisnpi
i.e. smallest n = 3
when it = -8
for z to be imaginary, z=|z|cis(npi+pi/2)
but kpi/3 <> npi+pi/2 for k and n integers :. never imaginary

for the 2nd, were you given that arg(z^2)=2argz?

 

[mojako]

z=2cis(pi/3) in mod-arg form
z^n = 2^n [cos (n*pi/3) + i sin (n*pi/3)]
for it to be real, imaginary part = 0
sin (n*pi/3) = 0
etc.

for imaginary,
cos(n*pi/3) = 0
n*pi/3 = -3pi/2, -pi/2, pi/2, 3pi/2, etc.
there wont be integral n


for the induction thing... I'll just prove the general case.
let z= r*cis@
let S(n) be defined by z^n = r^n*cis(n@), which satisfies both |z^n| = |z|^n and arg(z^n) = n*arg(z)
Clearly, by inspection(TM), S(1) is true.
If S(k) is true, z^k = r^k*cis(k@)
Consider the validity of S(k+1).
LHS
= (r*cis@)^(k+1)
= (r*cis@)^k * (r*cis@)
= r^k*cis(k@) * (r*cis@)
= r^(k+1) * [cis(k@)*cis@]
= r^(k+1) * (cosk@ + i sink@) (cos@ + i sin@)
= r^(k+1) * (cosk@cos@ + i cosk@sin@ + i sink@cos@ - sink@sin@)
= r^(k+1) * [cos(k+1)@ + i sin(k+1)@], by inspection(TM)
= RHS

So, whenever S(k), S(k+1) is true.
But S(1) is true.
Hence, S(n) is true for n=1,2,3,...

 

[blackfriday]

for the 2nd, were you given that arg(z^2)=2argz?

no

thanks for the help so far

as you can tell i only started 4u maths this week

 

[mojako]

no

thanks for the help so far

as you can tell i only started 4u maths this week

ur in the same situation with Estel

 

[Estel]

Your class moves quickly
We haven't even gotten up to square roots, let alone arguments =/

 

[nit]

Or you could do a little lemma thingy with IabI = IaIIbI, then do a nicer induction using just z's and k's...by inspection :uhhuh:

 

[mojako]

Or you could do a little lemma thingy with IabI = IaIIbI, then do a nicer induction using just z's and k's...by inspection :uhhuh:

u dont really need induction for that.
just say, "it follows that |z_1 z_2 ... z_n| = |z_1| + |z_2| + ... + |z_n|,
and when z_1=z_2=...=z_n, |z^n|=|z|^n"

and for the argument, find z*z
and you'll see that arg(z*z)=2arg(z)
then it follows that.....

Also if u find z*z then u have also proven that |ab|=|a||b| instead of assuming it.

now.. I dont understand.. why do we need to use induction to prove De Moivre's theorem? Isn't this convincing enough??