complex question (1 Viewer)

contro3ler · Nov 12, 2023

idont get this bro

Drongoski · Nov 12, 2023

Not as straightforward as I thought. Maybe there is a simple method.

Drongoski · Nov 12, 2023

z and w lie on the same line. Let:

$z = a cis\theta $ and $ w = b cis \theta \\ \\ \therefore acis\theta + bcis\theta = |a|bcis\theta\\ \\ \therefore (a+b)cis\theta = |a|bcis\theta\\ \\ \therefore a+b = |a|b \\ \\$

$|a+b| =||a|b| = |a|\times |b|$

Case 1) a>0 & b>0

$\therefore a + b = ab \implies ab-a = ab \implies a(b-1) = ab \implies a = \frac {b}{b-1}\\ \\ $i.e. $|z| = \frac {|w|}{|w|-1}$
Case 2) a>0 & b<0 and |a|>|b|

$a-b = a\times (-b) = -ab \implies a(b+1) = b \implies a = \frac {b}{b+1}\\ \\ $i.e. $ |z| = \frac {|w|}{|w| + 1}$

Case 3) a>0 & b<0 & |b| > |a|

$-b-a = a\times(-b) \implies ab-a = b \implies a = \frac {b}{b-1}\\ \\ $i.e. $|z| = \frac {-|w|}{-|w|-1} = \frac {|w|}{|w|+1}$

You can similarly consider the other 2 cases. There must be an easier way.

carrotsss · Nov 12, 2023

why is this thread glitching out so much

kendricklamarlover101 · Nov 12, 2023

Drongoski said:
z and w lie on the same line. Let:

$z = a cis\theta $ and $ w = b cis \theta \\ \\ \therefore acis\theta + bcis\theta = |a|bcis\theta\\ \\ \therefore (a+b)cis\theta = |a|bcis\theta\\ \\ \therefore a+b = |a|b \\ \\$

$|a+b| =||a|b| = |a|\times |b|$

Case 1) a>0 & b>0

$\therefore a + b = ab \implies ab-a = ab \implies a(b-1) = ab \implies a = \frac {b}{b-1}\\ \\ $i.e. $|z| = \frac {|w|}{|w|-1}$
Case 2) a>0 & b<0 and |a|>|b|

$a-b = a\times (-b) = -ab \implies a(b+1) = b \implies a = \frac {b}{b+1}\\ \\ $i.e. $ |z| = \frac {|w|}{|w| + 1}$

Case 3) a>0 & b<0 & |b| > |a|

$-b-a = a\times(-b) \implies ab-a = b \implies a = \frac {b}{b-1}\\ \\ $i.e. $|z| = \frac {-|w|}{-|w|-1} = \frac {|w|}{|w|+1}$

You can similarly consider the other 2 cases. There must be an easier way.

doesnt letting $z=acis\theta$ and $z=bcis\theta$ imply that a>0 and b>0 since they are the magnitudes of each complex number

fx82au · Nov 12, 2023

wow broken forum post

WeiWeiMan · Nov 12, 2023

carrotsss said:
why is this thread glitching out so much

I thought it was just me but on my laptop the thing got split in half

Nav123 · Nov 12, 2023

contro3ler said:
$Given that $z+w=|z|w$, where $|w|>1$, show that $|z|=\frac{|w|}{|w| \pm 1}$
View attachment 41492
idont get this bro

$z=w(|z|-1)$
$|z|=|w|||z|-1|$
$|z|^2=|w|^2||z|-1|^2$
$|z|^2=|w|^2(|z|^2-2|z|+1)$
$|z|^2(1-|w|^2)+2|w|^2|z|-|w|^2=0$

then just solve the quadratic for $|z|$

Drogonskis method is better and less algebra but this method requires less thinking

jimmysmith560 · Nov 12, 2023

It seems that the LaTeX function was incorrectly used in the original post, and this somehow caused this entire page to display incorrectly. I have removed the unnecessary LaTeX input (since the OP included a photo of his question anyway), which appears to have rectified the issue.

Life'sHard · Nov 13, 2023

Drop the nuclear latex codes

Drongoski · Nov 13, 2023

Nav123 said:
$z=w(|z|-1)$
$|z|=|w|||z|-1|$
$|z|^2=|w|^2||z|-1|^2$
$|z|^2=|w|^2(|z|^2-2|z|+1)$
$|z|^2(1-|w|^2)+2|w|^2|z|-|w|^2=0$

then just solve the quadratic for $|z|$

Drogonskis method is better and less algebra but this method requires less thinking

No. Your method is the type I was looking for. It's way better than Drongoski's plodding method.

complex question (1 Viewer)

contro3ler

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