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Sahithyan's S1
Sahithyan's S1 — Mathematics

Limits

limzz0f(z)=L\lim_{z\to z_0} f(z) = L iff:

ϵ>0  δ>0  z  (0<zz0<δ    f(z)L<ϵ)\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 limf(z)=L\lim f(z)=L.

  • limf(z)=L\lim \overline{f(z)}=\overline{L}
  • limRe(f(z))=Re(L)\lim \text{Re}(f(z))=\text{Re}(L)
  • limIm(f(z))=Im(L)\lim \text{Im}(f(z))=\text{Im}(L)

Real and imaginary limits

Let f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y), z0=x0+iy0z_0 = x_0 + iy_0, z=x+iyz=x+iy.

Suppose the real part and imaginary part limits to L1,L2L_1,L_2, which can be written as:

lim(x,y)(x0,y0)u(x,y)=L1      lim(x,y)(x0,y0)v(x,y)=L2\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:

limzz0f(z)=L1+iL2\lim_{z\to z_0} f(z)=L_1+iL_2

Difference from real functions

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 zz0\lvert z − z_0 \rvert decreases to 00.

Notes for questions

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

Important limits

limz0zz  doesn’t exist\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.

limz0zzz+z  doesn’t exist \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+tit+\sqrt{t}i.