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Basically I want to show the following:

$\sqrt{2}\ |z|\geq\ $|Re$z$| + |Im$z$|

So what I did is the following:

Let $z = a + bi$

Consider the following:

$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$

Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$

Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$

Hence $\sqrt{2}|z| \geq a + b$ but a is = |Re$z$|, b = |Im$z$|

Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.

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    Use Cauchy-Schwarz – 2017-01-13
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    Also, what you did is correct as well. – 2017-01-13
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    "Hence $\sqrt{2}|z| \geq a + b$ but a is = |Re$z$|, b = |Im$z$|". Note that $a=\text{Re}(z)$ and $b=\text{Im}(z)$. What you have to prove is $\sqrt 2\ |z|\ge |a|+|b|$ which you don't prove. – 2017-01-13
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    hmm, I'm abit confused. So how do I go about fixing the proof or should I try a different approach ? – 2017-01-13

4 Answers 4

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It suffices to consider $x, y > 0$. Other case simply changes the sign of $x,y$ and of $\theta$. Thus: $\dfrac{|\text{Re(z)}|}{|z|} = \dfrac{x}{\sqrt{x^2+y^2}}= \sin \theta, \dfrac{\text{Im(z)}}{|z|}= \dfrac{y}{\sqrt{x^2+y^2}} = \cos \theta, \theta \in \left(0,\dfrac{\pi}{2}\right)$. Thus you prove: $\sin \theta + \cos \theta \le \sqrt{2}$, which is true since $\sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)\le \sqrt{2}$ .

1

Let $z = r\cos\theta + ri\sin\theta$

Then $|Re(z)| + |Im (z)| = \pm\sqrt 2 r\sin(\theta \pm \frac{\pi}{4}) \leq \sqrt 2 r$

All four possible sign combinations will cover $\theta$ in all quadrants.

1

Let $z=a+bi$. Then $|z|^2=a^2+b^2$ and Re$z=a$ and Im$z=b$. Observe that $$(\sqrt{2}|z|)^2=2(a^2+b^2)$$ and $$\Big(|\text{Re}z|+|\text{Im}z|\Big)^2=a^2+2|ab|+b^2.$$

Now, $$ \begin{align} (|a|-|b|)^2\geq 0&\iff a^2+b^2-2|ab|\geq 0\\ &\iff2(a^2+b^2)\geq a^2+2|ab|+b^2\\ &\iff (\sqrt{2}|z|)^2\geq\Big(|\text{Re}z|+|\text{Im}z|\Big)^2\end{align} $$ Since $\sqrt{2}|z|$ and $|\text{Re}z|+|\text{Im}z|$ are nonnegative, the result follows.

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    Ah, I see where my mistakes are now. Thank you – 2017-01-13
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Observe \begin{align} |a|\cdot 1+|b|\cdot 1 \leq \sqrt{a^2+b^2}\sqrt{1^2+1^2}. \end{align}