I need a solution to this:
(problem attached by image)
P.S. 'z' is in the form z = x + iy
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.
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.
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 $\displaystyle \left|z\right|\leq 1$.