Question

Find the limits, if they exist;
i) lim z->0 Izl^2/z
ii) lim z->0 (z-re z)/im z

Answer 1

First, the definition of a complex limit is:
We write
\[ \lim_{z->a}f(z)=L, \]
If, for every \( \epsilon>0\) there exists a \( \delta > 0\) such that
\[ 0<|z-a|<\delta \Rightarrow |f(z)-L|<\epsilon \]
Using this, I claim the first limit is 0, since
\[ 0<|z-0|<\delta \Rightarrow \left|\frac{|z|^2}{z}-0\right|=|z|<\epsilon \]
When \( \epsilon=\delta. \)

The second is pretty obvious that it goes to 0. It's easily proved similarly to above.

Answer 2

Oops, meant that the second limit goes toward \( i \). Since
\[ z=Re(z)+Im(z)i \Rightarrow \frac{z-Re(z)}{Im(z)}=i \]
for all z.

Answer 3

can you also help me out on this question?
Let z belongs to C and a,b belong to R be arbitrary.
state whether the set is open, closed, or neither
i) 1< IzI<2
ii) im z >or=0 and 1<IzI<2

Answer 4

think i answered this one earlier. maybe someone else asked it

Answer 5

as for the second one, isn't
\[\frac{a+bi-a}{b}=i\]?

Answer 6

The first set is open, since all points are interior.
The second is neither, since the limit point 1.5 is included, but the limit point 2i is not.

Answer 7

or because the boundary is included on the real axis but not in the plane

Answer 8

interior of the plane rather.
one thing to be careful of however, is that \(\frac{|z^2|}{z}\neq |z|\) i think

Answer 9

re(z)=(1/2)(z+z)
im(z)=(1/2)(z-z)

how does [z-(re(z)]/im(z) =i ?

Answer 10

\[z=a+bi\] \(re(z)=a, im(z)=b, z-re(z)=a+bi-a=bi\)

Answer 11

and \(\frac{bi}{b}=i\) i think this is right

Answer 12

i got it. thanks

Answer 13

IzI^2/z is not equal to z. because lzl^2=a^2+b^2.
what is lzl^2/z =?

Answer 14

maybe easier if you write \(z=re^{i\theta}\)

Answer 15

I never claimed that |z|^2/z = z, I claimed that the absolute value of it was equal:
\[ \left| \frac{|z|^2}{z} \right|=\frac{|z|^2}{|z|} = |z| \]

Answer 16

ah yes, so you did, my fault

Answer 17

@satellite73
a<arg(z-z
state whether the set is open,closed or neither