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

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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.

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

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

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