Define the function f on the complex numbers by f (0) = 0 and 𝑓(z) =(𝑧̅)^2/𝑧 for z not equal to 0.?

Since 𝑧̅ and 𝑧 are reflections of one another over the real axis, we have that |𝑧̅| = |𝑧|.

Therefore, for all nonzero complex z:
|f(𝑧)| = |(𝑧̅)²/𝑧| = |𝑧̅|² / |𝑧| = |𝑧|² / |𝑧| = |𝑧|.

Since lim(𝑧→0) |𝑧| = 0, it follows that lim(𝑧→0) |f(𝑧)| ≤ 0 and thus
lim(𝑧→0) |f(𝑧)| = 0. This is true iff lim(𝑧→0) f(𝑧) = 0.

I hope this helps!

The limit exists=0.
Explanation:
Let z=x+yi, then
f(z)=[(x-yi)^2]/(x+yi)
=>
f(z)=Re(z)+Im(z)
=>
Re(z)=x(x^2-3y^2)/(x^2+y^2)
Im(z)=y(y^2-3x^2)/(x^2+y^2)
Let y=mx, where m is a real number.

Case 1: when m is finite, then
Re(z)=x(1-3m^2)/(1+m^2)->0 as x->0
Im(z)=mx(m^2-3)/(1+m^2)->0 as x->0
=>
f(z)->0, as x->0
since f(0)=0
=>limit f(z)=0
....z->0

Case 2: y/x=m, keeping x fixed, y->0
as m->0. Thus
Re(z)->x^3/x^2=x
Im(z)->0
&
f(z)->0 as y->0, then x->0
=>limit f(z)=0
x->0 y->0

Case 3: y/x=m, keeping y fixed, x->0
as m-> infinity. Thus
Re(z)->0 as x->0
Im(z)=(y^3)/y^2=y
&
f(z)->0 as x->0, then y->0
=>limit f(z)=0
y->0 x->0
So, no matter what the path, that x,y
chooses, is to tend to 0, f(z)->0. Hence
the limit exists.

f(z) = (z̅)² / z

f(0) = 0

Let z = x + iy → z̅ = x- iy

f(z) = (x - iy)² / (x + iy)

f(z) = x²-2xyi-y²/ (x + iy)

-x = 0 and y = 0

f(0) = 0.

| a + bi | = | a - bi|...thus | f(z)| = | z | ---> 0 as z ---> 0