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

Bobbie Comstock

Bobbie Comstock

Answered question

2022-01-16

How do I solve this problem: limz0Re(z2)z+Im(z)?

Answer & Explanation

Toni Scott

Toni Scott

Beginner2022-01-17Added 32 answers

Writing z=x+iy for some real x,y, we can rephrase this limit as
L=lim(x,y)(0,0)x2y2x+iy+y
This limit exists and equals zero. This follows once we show that
lim(x,y)(0,0)|x2y2x+iy+y|=lim(x,y)(0,0)|x2y2|(x+y)2+y2=0.
To this end, we use polar coordinates:
|x2y2|(x+y)2+y2=|r2cos(2θ)|(rcosθ+rsinθ)2+(rcosθ)2
=r|cos(2θ)|(cosθ+sinθ)2+cos2θ
Clearly |cos(2θ)| is bounded above by 1. Next, we need to show that (cosθ+sinθ)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
(cosθ+sinθ)2+cos2θ=(1+sin(2θ))+12(1+cos(2θ))
=32+12(2sin(2θ)+cos(2θ))
=32+125sin(2θ+α), for some α
32+12(5)
0.3820.
Hence, for any r0, there exists a constant C>0 such that

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