1. ## Complex Number problem

I need a solution to this:

(problem attached by image)

P.S. 'z' is in the form z = x + iy

2. Dear Bluedot,

I have attached the solution to this question. If you find the attachment unreadable or if you have any questions regarding this solution please reply to me.

Hope this helps.

This solution is wrong. Please don't refer it.

3. Originally Posted by Sudharaka
Dear Bluedot,

I have attached the solution to this question. If you find the attachment unreadable or if you have any questions regarding this solution please reply to me.

Hope this helps.
There are a few mistakes in your solution.

First, it is not necessarily correct that $|z| \geq |z-1|$: take, for example, $z = -1 \Rightarrow |z| = 1, ~ \text{but } |z-1| = |-2| = 2 > 1 = |z|$.

Second, the fact that $|arg z| \geq 0$ does not justify keeping the inequality sign. For example, if we have $a=2, b=3, c=\frac{1}{2}$ then $a \leq b$, but $a > bc = \frac{3}{2}$.

Third, I believe he was looking for a specific solution (finding for which z this inequality holds), rather than a proof that this statement is true.

4. Dear Defunkt,

Yes you are correct. Thank you for showing the mistakes of my solution. As you said he may want a solution for which the inequality is valid. Since I cannot delete the thread I have marked it wrong.

5. Third, I believe he was looking for a specific solution (finding for which z this inequality holds), rather than a proof that this statement is true.
I am looking for a proof to the statement for any value of z (not any specific solution).

6. If $\left|z\right|\geq 1$, then it is equivalent to prove that
$\left|\frac{1}{z}\right|+\left|1-\frac{1}{z}\right|\leq 1+\left|arg z\right|$
by letting $w=\frac{1}{z}$ , then $\left|w\right|\leq 1$, the above expression can be rewrited as
$\left|w\right|+\left|1-w\right|\leq 1+\left|arg w\right|$
This is obviously true from the view of geometry (length) and the triangle inequality.
similarly for the case $\left|z\right|\leq 1$.

7. Originally Posted by Shanks
similarly for the case $\left|z\right|\leq 1$.
If z = x + iy, where x and y are real, then |z| can never be less than 1, since z = sq.root(x^2+y^2)

(sorry, don't know how to use latex)

8. your claim is obviously false.
two simple counterexample: $z=0, \text{ or }z=\frac{1}{2}i$.

Can you show your effort on this prolem before asking help from other guys?