How do I solve this problem: limz→0Re(z2)z+Im(z)?

Writing ${\displaystyle z=x+iy}$ for some real x,y, we can rephrase this limit as
${\displaystyle L=\underset{(x,y)\to (0,0)}{lim}\frac{x^2-y^2}{x+iy+y}}$
This limit exists and equals zero. This follows once we show that
${\displaystyle \underset{(x,y)\to (0,0)}{lim}\left|\frac{x^2-y^2}{x+iy+y}\right|=\underset{(x,y)\to (0,0)}{lim}\frac{|x^2-y^2|}{\sqrt{{(x+y)}^2+y^2}}=0}$.
To this end, we use polar coordinates:
${\displaystyle \frac{|x^2-y^2|}{\sqrt{{(x+y)}^2+y^2}}=\frac{\left|r^2\cos \left(2\theta \right)\right|}{\sqrt{{(r\cos \theta +r\sin \theta )}^2+{\left(r\cos \theta \right)}^2}}}$
${\displaystyle =\frac{r\left|\cos \left(2\theta \right)\right|}{\sqrt{{(\cos \theta +\sin \theta )}^2+{\cos }^2\theta }}}$
Clearly ${\displaystyle \left|\cos \left(2\theta \right)\right|}$ is bounded above by 1. Next, we need to show that ${\displaystyle {(\cos \theta +\sin \theta )}^2+{\cos }^2\theta }$ 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
${\displaystyle {(\cos \theta +\sin \theta )}^2+{\cos }^2\theta =(1+\sin \left(2\theta \right))+\frac{1}{2}(1+\cos \left(2\theta \right))}$
${\displaystyle =\frac{3}{2}+\frac{1}{2}(2\sin \left(2\theta \right)+\cos \left(2\theta \right))}$
${\displaystyle =\frac{3}{2}+\frac{1}{2}\cdot \sqrt{5}\sin (2\theta +\alpha )}$, for some ${\displaystyle \alpha }$
${\displaystyle \ge \frac{3}{2}+\frac{1}{2}\cdot (-\sqrt{5})}$
${\displaystyle \approx 0.3820}$.
Hence, for any ${\displaystyle r\ge 0}$, there exists a constant ${\displaystyle C>0}$ such that

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