Question

Prove that for any complex number $z$,
$|\operatorname{Re} (z)| + |\operatorname{Im} (z)| \leqslant |z|\sqrt 2$
or $|x| + |y| \leqslant \sqrt 2 |x + iy|$

Answer 1

Hint:- $\operatorname{Re} (z)$is the real part of $z$ and $\operatorname{Im} (z)$ is the imaginary part of $z$.

If $z$ is any complex number then it can be written as,
$\Rightarrow z = x + iy = r{e^{i\theta }} = r(\cos \theta + i\sin \theta )$ (1)
As we know that $\operatorname{Re} (z) = x,\operatorname{Im} (z) = y{\text{ and }}|z| = r$
Using equation 1 we can write,
$\Rightarrow |x| + |y| = r[|\cos \theta | + |\sin \theta |]$ (2)
Now, to prove the given relation.
On squaring equation 2 we get,
$\Rightarrow {[|x| + |y|]^2} = {r^2}[{\cos ^2}\theta + {\sin ^2}\theta + |2\sin \theta \cos \theta |]$
As we know that, ${\cos ^2}\theta + {\sin ^2}\theta = 1$.
So, solving above equation it becomes,
$\Rightarrow {[|x| + |y|]^2} = {r^2}[1 + |2\sin \theta \cos \theta |]$
As we know that according to trigonometric identities,
$2\sin \theta \cos \theta = \sin 2\theta$
So, on solving the above equation. It becomes,
$\Rightarrow {[|x| + |y|]^2} = {r^2}[1 + |\sin 2\theta |]$
Now, as we know that, (3)
$\Rightarrow {\text{|}}\sin x| \leqslant 1$
So, the maximum value of ${\text{sin2}}\theta = 1$.
Now , equation 3 becomes,
$\Rightarrow {[|x| + |y|]^2} \leqslant 2{r^2}$
Now, taking square roots to both sides of the above equation. We get,
$\Rightarrow |x| + |y| \leqslant \sqrt 2 r$
So, above equation can be written as,
$\Rightarrow |\operatorname{Re} (z)| + |\operatorname{Im} (z)| \leqslant \sqrt 2 |z|$
Using equation 1 above equation can be written as,
$\Rightarrow |x| + |y| \leqslant \sqrt 2 |x + iy|$
Hence, $|\operatorname{Re} (z)| + |\operatorname{Im} (z)| \leqslant |z|\sqrt 2 {\text{ }}$or $|x| + |y| \leqslant \sqrt 2 |x + iy|$

Note:- Whenever you came up with this type of problem then easiest and efficient way
is to write complex number $z$, in polar form $\left( {z = r{e^{i\theta }}} \right)$, in terms of ${\text{sin}}\theta$and \cos \theta $$$$\left( {z = r(\cos \theta + i\sin \theta )} \right), or in terms of x and y $\left( {z = x + iy} \right)$ as per required result to be proved.