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

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Toni Scott · Accepted Answer

Writing z=x+iy for some real x,y, we can rephrase this limit as  L=lim(x,y)→(0,0)x2−y2x+iy+y  This limit exists and equals zero. This follows once we show that  lim(x,y)→(0,0)|x2−y2x+iy+y|=lim(x,y)→(0,0)|x2−y2|(x+y)2+y2=0.  To this end, we use polar coordinates:  |x2−y2|(x+y)2+y2=|r2cos⁡(2θ)|(rcos⁡θ+rsin⁡θ)2+(rcos⁡θ)2  =r|cos⁡(2θ)|(cos⁡θ+sin⁡θ)2+cos2θ  Clearly |cos⁡(2θ)| is bounded above by 1. Next, we need to show that (cos⁡θ+sin⁡θ)2+cos2θ 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  (cos⁡θ+sin⁡θ)2+cos2θ=(1+sin⁡(2θ))+12(1+cos⁡(2θ))  =32+12(2sin⁡(2θ)+cos⁡(2θ))  =32+12⋅5sin⁡(2θ+α), for some α  ≥32+12⋅(−5)  ≈0.3820.  Hence, for any r≥0, there exists a constant C&amp;gt;0 such that

How do I solve this problem: \lim_{z \rightarrow 0} \frac{Re(z^{2})}{z+Im(z)}?

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