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I want to prove

\begin{align} \lvert z\rvert &\geq \Re\{z\} \tag{1}\\ \lvert z\rvert &\geq \Im\{z\} \tag{2} \end{align}

I start with $z=x+iy$ so $$ \lvert z\rvert=\sqrt{x^2+y^2}\tag{3} $$

With the following (I guess it's valid for complex numbers?) $$ \lvert z\rvert =\sqrt{z^2} \iff \lvert z\rvert^2 =z^2 \tag{4} $$

I can write \begin{gather} \lvert z\rvert^2=x^2+y^2 \tag{5} \end{gather}

Using \begin{gather} \Re\{z\}=x \iff \Re\{z\}^2=x^2 \tag{6} \\ \Im\{z\}=y \iff \Im\{z\}^2=y^2 \tag{7} \end{gather} I can now write \begin{gather} \lvert z \rvert ^2=\Re\{z\}^2 +\Im\{z\}^2 \tag{8} \end{gather}

I'm stuck here.

What is the next step? Or should I stop here and conclude something from $(8)$?

Thanks!

Update: I not sure, but shouldn't we have absolute values in $(1)$ and $(2)$, i.e. $\lvert z\rvert \geq \lvert \Re\{z\}\rvert$ and $\lvert z\rvert \geq \lvert\Im\{z\}\rvert$?

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Hint:

Since $\Im \{z\}$ is a real number, you know that $\Im \{z\}^2 \geq 0$.

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  • $\begingroup$ Hi! Do you mean: Starting with $\lvert z \rvert ^2=\Re\{z\}^2 +\Im\{z\}^2$ we have $\lvert z \rvert ^2=(\Re\{z\}^2 +\Im\{z\}^2)\geq 0$. Therefore we kan remove $\Re\{z\}^2$ and because of this we can change the equality sign to an inequality sign, i.e. $\lvert z \rvert ^2\geq\Im\{z\}^2 \iff \lvert z \rvert \geq\Im\{z\}$? $\endgroup$ – JDoeDoe Feb 3 '18 at 10:56
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Note that

$$\begin{cases}\lvert z\rvert=\sqrt{x^2+y^2}\ge x\\\\\lvert z\rvert=\sqrt{x^2+y^2}\ge y\end{cases}$$

are always true.

Indeed they are true when $x,y<0$ and for $x,y\ge0$ we can square and obtain

$$\begin{cases}x^2+y^2\ge x^2\\\\x^2+y^2\ge y^2\end{cases}$$

As an alternative, note that in polar form the equations are equivalent to

$$\begin{cases}\rho\ge \rho \cos \theta \iff \cos \theta \le1\\\\\rho\ge \rho \sin \theta \iff\sin \theta \le1\end{cases}$$

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  • $\begingroup$ Hi! Do you mean: we start with $x^2\geq0 \iff x^2+y^2\geq y^2$ and with $\lvert z\rvert^2 =x^2+y^2$ we have: $\lvert z\rvert ^2\geq y^2 \iff $$\lvert z\rvert \geq y$. And finally with $\Im\{z\}=y$ we have $\lvert z\rvert\geq\Im\{z\}$. Is this algebraically correct? $\endgroup$ – JDoeDoe Feb 3 '18 at 13:29
  • $\begingroup$ Let consider $\lvert z\rvert=\sqrt{x^2+y^2}\ge x$, I mean that when $x<0$ the inequality is always satisfied since LHS is $\ge0$; when $x\ge0$ we are allowed to square the inequality without loss of information and obtain $x^2+y^2\ge x^2\implies y^2\ge0$ whiach is always true. By the way by the polar form it is trivially true. $\endgroup$ – user Feb 3 '18 at 13:33

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