Complex Numbers, EXT2 EQ-Bank 15

Prove that for any complex numbers \(z_1\) and \(z_2\),
\(\abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}\) (2 marks)
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Hence, or otherwise, show that if \(\abs{z-1}+\abs{z+1} \leqslant 4\) for \(z\in C,\)
\(\abs{z} \leqslant 2\) (2 marks)

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a. \(\text{Proof (See Worked Solutions)}\)

b. \(\text{See Worked Solutions}\)

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a. \(\text {Prove}\ \ \abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}:\)

\(\abs{z_1+z_2}^2\)	\(=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)\)
	\(=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2\)
	\(=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)\)

\(\text{Since}\ \ \operatorname{Re}(w) \leqslant\abs{w}\ \ \text{for} \ \ w\in C,\)

\(\abs{z_1+z_2}^2\)	\(\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}\)
	\(\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2\)
	\(\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2\)

\(\therefore\abs{z_1+z_2} \leqslant\abs{z_1}+\abs{z_2}\)

b. \(|z-1|+|z+1| \leqslant 4 \ \text{(given)}\ …\ (1)\)

\(\text {Using triangle inequality:}\)

\(|(z-1)+(z+1)| \leqslant|z-1|+|z+1|\)

\(\text{Since}\ \ (z-1)(z+1)=2 z:\)