Convergence of complex sequence $z_{n}$ A complex number is written in de form of $z=x+iy$, with $x,y \in \mathbb{R}$. We can call a sequence ($z_{n}$) of complex numbers convergent with limit $z^{*} \in \mathbb{C}$ if $\forall \varepsilon > 0: \exists n_{0} \in \mathbb{N} : \forall n \geq n_{0} : \vert z_{n} - z^{*} \vert < \varepsilon$.
Let $z_{n} = x_{n} + iy_{n}$ and $z^{*} =x^{*}+iy^{*}$, then $\vert z_{n} - z^{*} \vert = \sqrt{(x_{n} - x^{*})^{2} + (y_{n} - y^{*})^{2}}$.
$\textbf{Question:}$ Let $(z_{n})$ be a sequence of complex numbers. Take $x_{n}=$ Re $z_{n}$ and $y_{n} =$ Im $z_{n}$ for every $n \in \mathbb{N}$. Then $x_{n}$ and $y_{n}$ are both real sequences. Proof that $z_{n}$ is convergent if and only if the two sequences $x_{n}$ and $y_{n}$ are both convergent.
I started by proving the reverse of the theorem:
Suppose $x_{n}$ is convergent, so that $\lim_{n\to\infty} x_{n} = x^{*}$. Then $\forall \varepsilon > 0: \exists n_{1} \in \mathbb{N} : \forall n \geq n_{1} : \vert x_{n} - x^{*} \vert < \frac{\varepsilon}{\sqrt{2}}$. Also, suppose that $y_{n}$ is convergent, so that $\lim_{n\to\infty} y_{n} = y^{*}$. Then $\forall \varepsilon > 0: \exists n_{2} \in \mathbb{N} : \forall n \geq n_{2} : \vert y_{n} - y^{*} \vert < \frac{\varepsilon}{\sqrt{2}}$. Then, $\vert x_{n}-x^{*}\vert^{2}<\frac{\varepsilon^{2}}{2}$ and also $\vert y_{n}-y^{*}\vert^{2}<\frac{\varepsilon^{2}}{2}$. If we add both inequalities together, we get after some calculations $\sqrt{(x_{n}-x^{*})^{2}+(y_{n}-y^{*})^{2}}<\varepsilon$. This holds $\forall \varepsilon > 0$ and $\forall n \geq max \lbrace n_{1},n_{2}\rbrace$, by which follows that $z_{n}$ is convergent and $\lim_{n\to\infty} z_{n} = z^{*}$.
I'm not sure whether this proof is fully correct and I also have no idea how to proof the theorem the other way around.
 A: Assume $z_n \to z$. Hence for any $\varepsilon > 0$ we find $N \in \mathbb{N}$ such that $|z_n - z| < \varepsilon$ whenever $n > N$. Thus we get for any $n > N$ $$|\text{Re}z_n - \text{Re}z| = |\text{Re} (z_n - z)| \leq |z_n - z| < \varepsilon$$ and similarly for the imaginary part. Maybe for the other direction I would just use $|\cdot|$ rather than the definition with the squareroot for simplicity, i.e. assume $x_n \to x$ and $y_n \to y$, then we may show that $z_n \to z$ where $z_n := x_n + iy_n$ and $z := x + iy$. For any $\varepsilon > 0$ we find $N',N'' \in \mathbb{N}$ such that $$|x_n - x| < \varepsilon/2 \qquad \text{and} \qquad |y_n - y| < \varepsilon/2$$ whenever $n > N := \max\{N',N''\}$. Thus for any $n > N$ we have $$|z_n - z| = |x_n + iy_n - x - iy| \leq |x_n - x| + |y_n - y| < \varepsilon/2 + \varepsilon/2 = \varepsilon$$
A: $|x_n-x^*|< |z_n- z^*|$ so if the sequence $\{z_n\}$ converges and the rhs can be made arbitrarily small, so does the lhs and $\{x_n\}$ also converges. Same for $\{y_n\}$
Geometrically, you can think of $ |z_n- z^*| $ as the hypotenuse of a right triangle and of $|x_n-x^*|$ and $ |y_n- y^*|$ as the other sides, which are always smaller. Of course the inequalities can be proven by the monotone property of the square root and positivity of the squares involved
