FORM GR1268 Question 19

Question:
If \(z\) is a complex variable and \(\bar z\) denotes the complex conjugate of \(z\), which is \(\displaystyle\lim_{z\rightarrow0}\frac{(\bar z)^2}{z^2}\)?

(A) \(0\)     (B) \(1\)     (C) \(i\)     (D) \(\infty\)     (E) The limit does not exist.

Answer:
(E)

Answer Key:
Let \(z=x+iy\).
Then \(\bar z = x-iy\).

So, \(z^2=(x+iy)^2=x^2+2xyi+(iy)^2=x^2-y^2+2xyi\)
and \((\bar z)^2=(x-iy)^2=x^2-2xyi+(iy)^2=x^2-y^2-2xyi\).

Let \(z\) approach zero along the \(x\)-axis.
In other words, set \(y=0\).
Then
\[\lim_{x\rightarrow0}\frac{x^2-0^2-2x(0)i}{x^2-0^2+2x(0)i}
=\lim_{x\rightarrow0}\frac{x^2}{x^2}
=\lim_{x\rightarrow0}1=1.\]
Let \(z\) approach zero along the line \(y=x\).
In other words, set \(y=x\).
Then
\[\lim_{x\rightarrow0}\frac{x^2-x^2-2x^2i}{x^2-x^2+2x^2i}
=\lim_{x\rightarrow0}\frac{-2x^2i}{2x^2i}
=\lim_{x\rightarrow0}(-1)=-1.\]
Because the limits don't agree, the limit does not exist.

Alternatively, you can use polar coordinate.
Let \(z=re^{i\theta}\).
Then \(\bar z=re^{-i\theta}\).
\[\lim_{r\rightarrow0}\frac{(\bar z )^2}{z^2}
=\lim_{r\rightarrow0}\frac{r^2e^{-i2\theta}}{r^2e^{i2\theta}}
=\lim_{r\rightarrow0}e^{-i4\theta}
=e^{-i4\theta}.\]
The limit does not exist because it depends on \(\theta\).

\(z\) approaching zero along \(x\)-axis corresponds to \(\theta=0\),
in which case, the limit is \(e^{-i4(0)}=e^0=1\).

\(z\) approaching zero along the line \(y=x\) corresponds to \(\theta=\frac{\pi}{4}\),
in which case, the limit is \(e^{-i4\left(\frac{\pi}{4}\right)}=e^{-\pi i}=-1\).