Theorem: convergence of complex sequence

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Homework Statement



show that a sequence zn (of complex numbers) converges if and only if it's real and imaginary parts converge.

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The Attempt at a Solution



if the real parts converge and the imaginary parts converge then using linearity, we have:
lim(real + imaginary)= lim(real) + lim(imaginary)

I think this is a wrong proof (and surely incomplete). does anyone have any ideas?
 
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hi tinytim,
thanks :smile:,
ok so here it goes:
we have 'if and only if' so first I am going to 'try' and prove it in the first direction:
let zn be a convergent sequence (and converges to zc), then:
for all [tex]\epsilon>0[/tex] there exists [tex]\delta[/tex] such that [tex]\left|z_n-z_c\right|<\epsilon[/tex] for all n>[tex]\delta[/tex].
let zc=a+ib:
[tex]\left|z_n-(a+ib)\right|=\left|z_n+(-a-ib)\right|\leq\left|z_n\right|+\left|(-a-ib)\right|=sqrt(a^2+b^2)[/tex]
(i used the triangle inequality)
but now i don't know what to do.
 
Think about this:

If [tex]z_1 =e + f i[/tex] and [tex]z_2 = g + h i[/tex] are complex numbers, then

[tex] |e - g | \le |z_1 - z_2|[/tex]

and

[tex] |f - h| \le |z_1 - z_2|[/tex]
 
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i agree, but why would this help here, do i need to use this to put
[tex]\left|z_n\right|+\sqrt{a^2+b^2}[/tex] in terms of something that could be my [tex]\epsilon[/tex]
 
Suppose you want to show that

[tex] z_n = x_n + y_ni \to z = x+yi[/tex]
if, and only if

[tex] x_n \to x \text{ and } y_n \to y[/tex]

If you know the complex sequence converges then the inequalities will show the two real sequences converge. If the real sequences converge, you can show the complex one does as well.
 
oh right i see,
so first, to prove in the first direction:
let zn be a convergent sequence, then
lim(zn)=lim(xn)+i*lim(yn)

zn-zc=(xn-x)+ i (yn-y)
where zc=(x,y) is the limit of the sequence zn.
so:
[tex]\left|x_n-x\right|\leq\left|z_n-z_c\right|[/tex] and [tex]\left|y_n-y\right|\leq\left|z_n-z_c\right|[/tex]
is this right? buti don't fully understand why these ineqaulities are true?