# Complex Number problem

• Jan 4th 2010, 12:17 AM
Bluedot
Complex 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 AM
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.

This solution is wrong. Please don't refer it.
• Jan 4th 2010, 10:49 AM
Defunkt
Quote:

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 $\displaystyle |z| \geq |z-1|$: 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 PM
Sudharaka
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 AM
Bluedot
Quote:

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).
• Jan 5th 2010, 10:51 PM
Shanks
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 AM
Bluedot
Quote:

Originally Posted by Shanks
similarly for the case $\displaystyle \left|z\right|\leq 1$.

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)
• Jan 6th 2010, 03:56 AM
Shanks
two simple counterexample: $\displaystyle z=0, \text{ or }z=\frac{1}{2}i$.