complex no. modulus

**afeasfaerw23231233** · May 18th 2008, 10:28 PM

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

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

**Moo** · May 18th 2008, 11:56 PM

Hello,

Originally Posted by afeasfaerw23231233

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

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

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

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

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

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

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

**Moo** · May 19th 2008, 12:03 AM

Originally Posted by afeasfaerw23231233

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

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

[snip]

Here, your working will help

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

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

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

What can you conclude ?

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

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

**afeasfaerw23231233** · May 19th 2008, 10:33 AM

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

**Moo** · May 19th 2008, 10:43 AM

Originally Posted by afeasfaerw23231233

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

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

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

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