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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.

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.

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