Need help quick! (1 Viewer)

[currysauce]

I have questions i don't understand

Use the properties of modulus and argument of a complex number to deduce that

a) conj (z1z2) = conj (z1) times conj (z2)

b) conj ( 1 / z ) = 1 / (conj z)

c) conj ( z1 / z2 ) = (conj z1) / (conj z2)




2. Use the method of mathematical induction to prove that | z^n| = |z| ^n and arg(z^n) = n arg z for all postitive integers n.

AHHHH

 

[VieTz_88]

Umm for 1 just let z1 = a+ib, z2= c+id
therefore, z1z2 = (a+ib)(c+id)
= ac-bd+i(ad+bc)
therefore, conj(z1z2) = ac-bd-i(ad+bc)
Also, conj(z1).conj(z2) = (a-ib)(c-id)
= ac+bd-i(ad+bc)
= conj(z1z2)
You do the same for (b) and (c) .... let z1 = a+ib, z2= c+id :uhhuh:

 

[Slidey]

I shall henceforth denote the conjugate of z as z_bar:

"a) conj (z1z2) = conj (z1) times conj (z2)

b) conj ( 1 / z ) = 1 / (conj z)

c) conj ( z1 / z2 ) = (conj z1) / (conj z2"

a:
Given (z₁.z₂)_bar
Want (z₁)_bar * (z₂)_bar
Such that (z₁.z₂)_bar = (z₁)_bar * (z₂)_bar
Let z₁=x+iy, z₂=a+ib

LHS = ([x+iy].[a+ib])_bar
([x+iy].[a+ib])_bar = (ax-by+[ay+bx].i)_bar
(ax-by+[ay+bx].i)_bar = ax-by-(ay+bx)i
RHS = (x-iy) * (a-ib) = ax-by-(ay+bx)i
LHS=RHS, Q.E.D.

b:
Given (1/z)_bar
Want 1/(z_bar)
Such that (1/z)_bar = 1/(z_bar)
Let z=x+iy

LHS = [1/(x+iy)]_bar = [(x-iy)/(x^2+y^2)]_bar = (x+iy)/(x^2+y^2)
RHS = 1/(x-iy) = (x+iy)/(x^2+y^2) = LHS
Q.E.D.

c:
Given (z₁/z₂)_bar
Want z₁_bar/z₂_bar
Such that (z₁/z₂)_bar = z₁_bar/z₂_bar
Let z₁=x+iy, z₂=a+ib

LHS = [(x+iy)/(a+ib)]_bar = [(x+iy)(a-ib)/(a^2+b^2)]_bar = (x-iy)(a+ib)/(a^2+b^2) {from proofs a and b}
RHS = (x-iy)/(a-ib) = (x-iy)(a+ib)/(a^2+b^2) = LHS
Q.E.D.

 

[Slidey]

2. Use the method of mathematical induction to prove that | z^n| = |z| ^n and arg(z^n) = n arg z for all postitive integers n.

I'll do both at once, OK?

We want that P(n)= z^n=r^n.cis(n@), where r=|z| and @=Arg(z), cis@=(cos@+isin@)

Now for n=1,
P(1)=z^1=r^1cis(1.@)=rcis@, which is true, so n=1 is true
Let n=k, assume it is true for for P(k):
z^k=r^k.cis(k.@)
We want that P(k+1) is true if P(k) is true:
P(k+1)=r^(k+1).cis([k+1].@)

Now, z=rcis@ (true), z^k=cis(k@) (assumed true).
z^k.z=(r^n.cis[n@])(rcis@)=r^n.r.cis(n@+@)
z^(k+1)=r^(k+1).cis([k+1]@)
Hence if it is true for P(k), it is true for P(k+1).
It is true for P(1), and hence for P(2), P(3) and so on.
So P(n)=z^n=r^n.cis(n@) is true for all n>=1.

Now from this, |z^n|=r^n, but r=|z|, so r=|z|^n=|z^n|
Also from this, arg(z^n)=n@, but @=arg(z), so arg(z^n) = n.arg(z)