Complex numbers from cambridge (1 Viewer)

hanod · Oct 15, 2022

Can some one show me the solution for the following question from Cambridge book

let z1,z2 and z3 be three complex numbers with modulus 1. prove that |z1z2+z1z3+z2z3| = |z1+z2+z3|

Lith_30 · Oct 15, 2022

using the identity $z\bar{z}=|z|^2$

$\begin{align*}\text{LHS}^2&=|z_1z_2+z_1z_3+z_2z_3|^2\\&=(z_1z_2+z_1z_3+z_2z_3)\overline{(z_1z_2+z_1z_3+z_2z_3)}\\&=|z_1|^2|z_2|^2+|z_1|^2z_2\bar{z_3}+z_1|z_2|^2\bar{z_2}+|z_1|^2\bar{z_2}z_3+|z_1|^2|z_3|^2+z_1\bar{z_2}|z_3|^2+\bar{z_1}|z_2|^2z_3+\bar {z_1}z_2|z_3|^2+|z_2|^2|z_3|^2\\&=|z_1|^2+z_2\bar{z_3}+z_1\bar{z_2}+\bar{z_2}z_3+|z_3|^2+z_1\bar{z_2}+\bar{z_1}z_3+\bar {z_1}z_2+|z_2|^2\\&=z_1\bar{z_1}+z_2\bar{z_3}+z_1\bar{z_2}+\bar{z_2}z_3+z_3\bar{z_3}+z_1\bar{z_2}+\bar{z_1}z_3+\bar {z_1}z_2+z_2\bar{z_2}\end{align*}$

which will eventually simplify to $(z_1+z_2+z_3)\overline{(z_1+z_2+z_3)}$ which is $RHS^2$

hanod · Oct 15, 2022

Lith_30 said:
using the identity $z\bar{z}=|z|^2$

$\begin{align*}\text{LHS}^2&=|z_1z_2+z_1z_3+z_2z_3|^2\\&=(z_1z_2+z_1z_3+z_2z_3)\overline{(z_1z_2+z_1z_3+z_2z_3)}\\&=|z_1|^2|z_2|^2+|z_1|^2z_2\bar{z_3}+z_1|z_2|^2\bar{z_2}+|z_1|^2\bar{z_2}z_3+|z_1|^2|z_3|^2+z_1\bar{z_2}|z_3|^2+\bar{z_1}|z_2|^2z_3+\bar {z_1}z_2|z_3|^2+|z_2|^2|z_3|^2\\&=|z_1|^2+z_2\bar{z_3}+z_1\bar{z_2}+\bar{z_2}z_3+|z_3|^2+z_1\bar{z_2}+\bar{z_1}z_3+\bar {z_1}z_2+|z_2|^2\\&=z_1\bar{z_1}+z_2\bar{z_3}+z_1\bar{z_2}+\bar{z_2}z_3+z_3\bar{z_3}+z_1\bar{z_2}+\bar{z_1}z_3+\bar {z_1}z_2+z_2\bar{z_2}\end{align*}$

which will eventually simplify to $(z_1+z_2+z_3)\overline{(z_1+z_2+z_3)}$ which is $RHS^2$

Big thank you

hanod · Oct 15, 2022

hogzillaAnson said:
Call them a,b,c instead for simplicity. A few things of note:
$z = x+yi,\ x,y \in \mathbb{R} \implies z\overline{z} = (x+yi)(x-yi) = x^2 + y^2 =|z|^2$
Also,
$\text{If } z \text{ lies on the unit circle, then}\ |z| = z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$
So,
$\begin{align*}\operatorname{LHS}^2= |a+b+c|^2 &= (a+b+c)\overline{(a+b+c)} = (a+b+c)(\overline a+\overline b+ \overline c) = (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \\ &= 3 + \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} \\ \operatorname{RHS}^2 = |ab+bc+ca|^2 &= (ab+bc+ca)\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = 3 + \frac{ab}{bc} + \frac{bc}{ab} + \frac{bc}{ca} + \frac{ca}{bc} + \frac{ab}{ca} + \frac{ca}{ab} \\ &= 3 + \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} = \operatorname{LHS}^2 \\ \therefore \operatorname{LHS} &= \operatorname{RHS}\ [\operatorname{LHS}, \operatorname{RHS} \geq 0]\end{align*}$
To be honest, we don't need to use the fact that the conjugate is the reciprocal here, but it just made typesetting easier. The key is to use the fact that

$|z|^2 = z\overline{z}\ \ \forall z \in \mathbb{C}$

then expand and equate

Big thank you

idkkdi · Oct 15, 2022

hanod said:
Can some one show me the solution for the following question from Cambridge book

let z1,z2 and z3 be three complex numbers with modulus 1. prove that |z1z2+z1z3+z2z3| = |z1+z2+z3|

|z1+z2+z3| = |conj(z1+z2+z3)|
Since mod 1, conj is 1/z_n, |z1+z2+z3| = |1/z1 + 1/z2 + 1/z3|
Multiply RHS by |z1||z2||z3| = 1,
|z1+z2+z3| = |z1z2+z1z3+z2z3|

Complex numbers from cambridge (1 Viewer)

hanod

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Lith_30

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hanod

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hanod

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idkkdi

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