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Now, I know this question has been asked here but my question doesn't deal with finding a solution, my question deals with checking the validity of the question.


Question:- If $z_1, z_2$ are complex numbers and $c$ is a positive number. Prove that $$|z_1-z_2|^2 \lt (1+c)|z_1|^2+\left(1+\frac{1}{c} \right)|z_2|^2$$


My solution:-

As, $|z_1+z_2|^2=|z_1|^2+|z_2|^2+2|z_1||z_2|\cos{\theta} \text{, where $\theta $ is the angle between $z_1$ and $z_2$}$

Now, $$\begin{equation} \dfrac{c|z_1|^2+\dfrac{1}{c}|z_2|^2}{2} \ge \sqrt{c|z_1|^2\cdot\dfrac{1}{c}|z_2|^2}=|z_1||z_2| \\ \implies c|z_1|^2+\dfrac{1}{c}|z_2|^2 \ge 2|z_1||z_2|\ge 2|z_1||z_2|\cos{\theta} \\ \implies c|z_1|^2+\dfrac{1}{c}|z_2|^2 \ge 2|z_1||z_2|\cos{\theta} \\ \implies (1+c)|z_1|^2+\left(1+\dfrac{1}{c}\right)|z_2|^2 \ge |z_1|^2+|z_2|^2+2|z_1||z_2|\cos{\theta} \end{equation}$$

Now, to check whether the equality holds , we see whether the condition $c|z_1|^2=\dfrac{1}{c}|z_2|^2$ holds, which does hold if $|z_1|=|z_2|=0$


My deal with the question:- Why didn't the question include the equality case in the question, or is it something that I did wrong, and if you have a more elegant solution please do provide it.

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  • $\begingroup$ The equality case should be included, i.e. we should have $\leq$ instead of just $<$, because clearly if $z_1=z_2=0$ we have equality. $\endgroup$
    – M10687
    Jul 17 '16 at 20:57
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This answer assumes that you meant $|z_1\color{red}{+}z_2|^2$ instead of $|z_1-z_2|^2$.

Why didn't the question include the equality case in the question

The claim in the question is false. As M10687 comments, for $z_1=z_2=0$, the inequality does not hold.

is it something that I did wrong

If you want to prove that

$$|z_1\color{red}{+}z_2|^2 \color{red}{\le} (1+c)|z_1|^2+\left(1+\frac{1}{c} \right)|z_2|^2$$ for $z_1, z_2\in\mathbb C$ and $c\gt 0$, then your solution is fine. You did nothing wrong.

if you have a more elegant solution please do provide it.

You may want to see the hints in the answers in the link you provided.

Using Matt L.'s hint,

$$\begin{align}&(1+c)|z_1|^2+\left(1+\frac 1c\right)|z_2|^2-|z_1+z_2|^2\\&=(1+c)z_1\overline{z_1}+\left(1+\frac 1c\right)z_2\overline{z_2}-(z_1+z_2)(\overline{z_1}+\overline{z_2})\\&=cz_1\overline{z_1}+\frac 1cz_2\overline{z_2}-z_1\overline{z_2}-\overline{z_1}z_2\\&=cz_1\overline{z_1}+\frac 1cz_2\overline{z_2}-\sqrt c\cdot\frac{1}{\sqrt c}z_1\overline{z_2}-\sqrt c\cdot\frac{1}{\sqrt c}\overline{z_1}z_2\\&=\left(\sqrt c\ z_1-\frac{1}{\sqrt c}z_2\right)\left(\sqrt c\ \overline{z_1}-\frac{1}{\sqrt{c}}\overline{z_2}\right)\\&=\left|\sqrt c\ z_1-\frac{1}{\sqrt c}z_2\right|^2\\&\ge 0\end{align}$$

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