I need a solution to this:

(problem attached by image)

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

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- Jan 4th 2010, 12:17 AMBluedotComplex Number problem
I need a solution to this:

(problem attached by image)

P.S. 'z' is in the form z = x + iy - Jan 4th 2010, 07:49 AMSudharaka
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.** - Jan 4th 2010, 10:49 AMDefunkt
There are a few mistakes in your solution.

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

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

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. - Jan 4th 2010, 06:50 PMSudharaka
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. - Jan 5th 2010, 07:35 AMBluedotQuote:

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.

- Jan 5th 2010, 10:51 PMShanks
If $\displaystyle \left|z\right|\geq 1$, then it is equivalent to prove that

$\displaystyle \left|\frac{1}{z}\right|+\left|1-\frac{1}{z}\right|\leq 1+\left|arg z\right|$

by letting $\displaystyle w=\frac{1}{z}$ , then $\displaystyle \left|w\right|\leq 1$, the above expression can be rewrited as

$\displaystyle \left|w\right|+\left|1-w\right|\leq 1+\left|arg w\right|$

This is obviously true from the view of geometry (length) and the triangle inequality.

similarly for the case $\displaystyle \left|z\right|\leq 1$. - Jan 6th 2010, 03:42 AMBluedot
- Jan 6th 2010, 03:56 AMShanks
your claim is obviously false.

two simple counterexample: $\displaystyle z=0, \text{ or }z=\frac{1}{2}i$.

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