Bobbie Comstock

2022-01-16

How do I solve this problem: $\underset{z\to 0}{lim}\frac{Re\left({z}^{2}\right)}{z+Im\left(z\right)}$?

Toni Scott

Writing $z=x+iy$ for some real x,y, we can rephrase this limit as
$L=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{x+iy+y}$
This limit exists and equals zero. This follows once we show that
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}|\frac{{x}^{2}-{y}^{2}}{x+iy+y}|=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{|{x}^{2}-{y}^{2}|}{\sqrt{{\left(x+y\right)}^{2}+{y}^{2}}}=0$.
To this end, we use polar coordinates:
$\frac{|{x}^{2}-{y}^{2}|}{\sqrt{{\left(x+y\right)}^{2}+{y}^{2}}}=\frac{|{r}^{2}\mathrm{cos}\left(2\theta \right)|}{\sqrt{{\left(r\mathrm{cos}\theta +r\mathrm{sin}\theta \right)}^{2}+{\left(r\mathrm{cos}\theta \right)}^{2}}}$
$=\frac{r|\mathrm{cos}\left(2\theta \right)|}{\sqrt{{\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}^{2}+{\mathrm{cos}}^{2}\theta }}$
Clearly $|\mathrm{cos}\left(2\theta \right)|$ is bounded above by 1. Next, we need to show that ${\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}^{2}+{\mathrm{cos}}^{2}\theta$ attains a positive minimum value; this assures that the trigonometric factor in the limit above is bounded (no division by zero).
To this end, observe that
${\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}^{2}+{\mathrm{cos}}^{2}\theta =\left(1+\mathrm{sin}\left(2\theta \right)\right)+\frac{1}{2}\left(1+\mathrm{cos}\left(2\theta \right)\right)$
$=\frac{3}{2}+\frac{1}{2}\left(2\mathrm{sin}\left(2\theta \right)+\mathrm{cos}\left(2\theta \right)\right)$
$=\frac{3}{2}+\frac{1}{2}\cdot \sqrt{5}\mathrm{sin}\left(2\theta +\alpha \right)$, for some $\alpha$
$\ge \frac{3}{2}+\frac{1}{2}\cdot \left(-\sqrt{5}\right)$
$\approx 0.3820$.
Hence, for any $r\ge 0$, there exists a constant $C>0$ such that

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