Thread: complex no. modulus

p289 q20
question: For any complex number z, let $\displaystyle \bar z$ and |z| be its conjugate and modulus respectively. suppose z_1 and z_2 are complex numbers such that $\displaystyle |z_1|=|z_2|=|z_1+z_2|=1$
(a)By using the property $\displaystyle z \bar z = |z|^2$, find the value of $\displaystyle (z_1+z_2) (\frac 1 z_1 + \frac 1 z_2)$
(b)Using the result of (a) , show that $\displaystyle z_1 ^2+z_1 z_2 +z_2^2 = 0$. Hence deduce that $\displaystyle z_1^3=z_2^3$

my working:
$\displaystyle =\frac {(z_1+z_2)^2} {z_1 z_2}$
$\displaystyle =\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}$
$\displaystyle = 2 +\frac {z_1}{ z_2} + \frac{ z_2 }{z_1}$
stuck here and don't know how to do. thanks!

Hello,

Originally Posted by afeasfaerw23231233

p289 q20
question: For any complex number z, let $\displaystyle \bar z$ and |z| be its conjugate and modulus respectively. suppose z_1 and z_2 are complex numbers such that $\displaystyle |z_1|=|z_2|=|z_1+z_2|=1$
(a)By using the property $\displaystyle z \bar z = |z|^2$, find the value of $\displaystyle (z_1+z_2) (\frac 1 z_1 + \frac 1 z_2)$

Remember that generally, if you have a complex number in a denominator, multiply by its conjugate

$\displaystyle \frac{1}{z_1}=\frac{\bar z_1}{z_1 \bar z_1}=\frac{\bar z_1}{|z_1|^2}=\boxed{\bar z_1}$

The same way, we get :
$\displaystyle \boxed{\frac{1}{z_2}=\bar z_2}$

$\displaystyle \implies (z_1+z_2)\left(\frac{1}{z_1}+\frac{1}{z_2}\right)= (z_1+z_2)(\bar z_1+\bar z_2)$

Noticing that $\displaystyle \bar z_1+\bar z_2$ is the conjugate of $\displaystyle z_1+z_2$, what can you conclude ?


(it's equal to 1 - I put it because we need it for the following)

Originally Posted by afeasfaerw23231233

(b)Using the result of (a) , show that $\displaystyle z_1 ^2+z_1 z_2 +z_2^2 = 0$.

my working:
$\displaystyle =\frac {(z_1+z_2)^2} {z_1 z_2}$
$\displaystyle =\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}$

[snip]

Here, your working will help

$\displaystyle 1=(z_1+z_2)\left(\frac{1}{z_1}+\frac{1}{z_2}\right )=\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}$
This is what you've written.

$\displaystyle \begin{aligned}1 &=\frac{z_1 z_2+{\color{blue}z_1^2+z_2^2+z_1 z_2}}{z_1 z_2} \\
{\color{red}1} &={\color{red}1}+\frac{{\color{blue}z_1^2+z_2^2+z_ 1 z_2}}{z_1 z_2} \end{aligned}$

$\displaystyle \implies \frac{{\color{blue}z_1^2+z_2^2+z_1 z_2}}{z_1 z_2}=0$

What can you conclude ?

Hence deduce that $\displaystyle z_1^3=z_2^3$

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

thank you very much! next time can i treat $\displaystyle \frac 1 z = \bar z$ as an identity and write it directly ?

Originally Posted by afeasfaerw23231233

thank you very much! next time can i treat $\displaystyle \frac 1 z = \bar z$ as an identity and write it directly ?

No, it's only true if $\displaystyle |z|=1$ ...

But in general, when you have a complex number in the denominator, multiply it by its conjugate, because it'll always yield a real number, and it's easier to deal with it