Limits

$\lim_{z\to z_0} f(z) = L$ iff:

\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{z}\; \big(0<|z-z_0|<\delta\implies{|f(z)-L|<\epsilon})

Properties

All properties mentioned in Limits | Real Analysis are applicable to complex limits. Additional properties are mentioned below:

Suppose $\lim f(z)=L$ .

Let $f(z)=u(x,y)+iv(x,y)$ , $z_0 = x_0 + iy_0$ , $z=x+iy$ .

Suppose the real part and imaginary part limits to $L_1,L_2$ , which can be written as:

\lim_{(x,y)\to(x_0,y_0)} u(x,y)=L_1 \;\;\; \lim_{(x,y)\to(x_0,y_0)} v(x,y)=L_2

Then:

\lim_{z\to z_0} f(z)=L_1+iL_2

For real functions, when considering the limit at a point, the limit could be be approaching the point either from left or right.

For complex functions, the point can be approached along any path in the complex plane. The distance $\lvert z - z_0 \rvert$ decreases to $0$ .

When 2 arbitrary paths are chosen: if the limits on each are different, then the limit DNE.
When substituting $z=x+imx$ , if $m$ doesn’t cancel out, then the limit DNE.
In most limits, subtituting $z=re^{i\theta}$ will simplify the limit a lot.
In very complex functions, limits can be taken for real and imaginary parts separately.

\lim_{z\to 0} \frac{z}{\overline{z}}\;\text{doesn't exist}

The above limit is important as it shows up in many questions. Can be disproved by taking two paths: real, imaginary axes.

\lim_{z\to 0} \frac{z \overline{z}}{z+\overline{z}}\;\text{doesn't exist}

Can be proven usign taking 2 paths: real axis, $t+\sqrt{t}i$ .

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October 17, 2026 1 min read

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