1. ## Question(s) about complex numbers.

• I understand that $|z| = \sqrt{x^2+y^2}$ (from the Argand Diagram). Is there an algebraic way to show that $|z| = \sqrt{x^2+y^2}$?

• Why is the principal value of the argument taken to be between $-\pi$ and $\pi$? I was asked to find the modulus and argument of $-1-i$. Here is what I did:

$|z| = \sqrt{\left(-1\right)^2+\left(-1\right)^2} = \sqrt{1+1} = \sqrt{2}$. Right.

$\arg{z} = \arctan\left(\frac{-1}{-1}\right)+\frac{\pi}{2} = \arctan(1)+\frac{\pi}{2} = \frac{\pi}{4}+2\pi = \frac{2\pi+\pi}{4} = \frac{3\pi}{4}$. Obviously that isn't between $-\pi$ & $\pi$. What am I supposed to do then?

PS. For the last part (as noticed by Archie Meade), I meant: $\arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4}$.

2. $There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$

$The\ angle\ is\ taken\ as\ 0\ to\ \pi\ radians\ if\ z\ is\ on\ or\ above\ the\ horizontal\ axis.$

$It's\ taken\ as\ 0\ to\ -\pi\ for\ z\ below\ the\ horizontal\ axis.$
$It's\ the\ convention.$

$0\ to\ -\pi\ are\ the\ angles\ from\ 360\ to\ 180\ degrees.$

$Arctan\ doesn't\ distinguish\ between\ opposite\ ends\ of\ the\ same\ line,$
$so\ you\ should\ ask\ yourself\ why\ you\ are\ adding\ 90\ degrees.$

$Adding\ \frac{\pi}{2}\ radians\ rotates\ a\ line\ through\ 90\ degrees.$

$From\ the\ real\ and\ imaginary\ parts\ of\ the\ complex\ number,$
$you\ can\ identify\ it\ as\ being\ between\ 180\ and\ 270\ degrees.$

$You\ must\ make\ this\ evaluation\ first,\ then\ calculate$
$the\ acute\ angle\ z\ makes\ with\ either\ the\ real\ or\ imaginary\ axis.$

3. Here is the way to find the principal argument.
$\arg (x + yi) = \left\{ {\begin{array}{rl}
{\arctan (y/x)} & {,0 < x} \\
{\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0} \\
{\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0} \\

\end{array} } \right.$

4. Originally Posted by Archie Meade
$There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$
I understand one (using pythagoras' theorem). What is the other? And is there any other way to show it apart from plotting the Argand Diagram?
$The\ angle\ is\ taken\ as\ 0\ to\ \pi\ radians\ if\ z\ is\ on\ or\ above\ the\ horizontal\ axis.$
$It's\ taken\ as\ 0\ to\ -\pi\ for\ z\ below\ the\ horizontal\ axis.$
$It's\ the\ convention.$ $0\ to\ -\pi\ are\ the\ angles\ from\ 360\ to\ 180\ degrees.$
Thank you. That is what I have been missing.
$Arctan\ doesn't\ distinguish\ between\ opposite\ ends\ of\ the\ same\ line,$
$so\ you\ should\ ask\ yourself\ why\ you\ are\ adding\ 90\ degrees.$
$Adding\ \frac{\pi}{2}\ radians\ rotates\ a\ line\ through\ 90\ degrees.$
That was a mistake. What I meant was $\arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4}$
$From\ the\ real\ and\ imaginary\ parts\ of\ the\ complex\ number,$
$you\ can\ identify\ it\ as\ being\ between\ 180\ and\ 270\ degrees.$
I could see that. But I was measuring anticlockwise from zero. I thought that the angle I needed was the acute angle I found plus 180 (measuring anticlockwise from zero).

$You\ must\ make\ this\ evaluation\ first,\ then\ calculate$ $the\ acute\ angle\ z\ makes\ with\ either\ the\ real\ or\ imaginary\ axis.$
Alright. I will do that. Little me first take a little grasp of what I've so far gathered.

5. Originally Posted by Plato
Here is the way to find the principal argument.
$\arg (x + yi) = \left\{ {\begin{array}{rl}
{\arctan (y/x)} & {,0 < x} \\
{\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0} \\
{\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0} \\

\end{array} } \right.$
So for $-1-i$, $\arg(z) = \arctan\left(\frac{-1}{-1}\right)-\pi = \arctan\left(1\right)-\pi = \frac{\pi}{4}-\pi = \frac{\pi-4\pi}{4} = -\frac{3\pi}{4}$. Brilliant. Thanks.

6. $Hi\ Captcha,$

$The\ proof\ is\ just\ 2\ alternative\ ways\ to\ derive\ Pythagoras'\ Theorem.$

$One\ way\ is\ to\ split\ a\ right-angled\ triangle\ into\ two$
$smaller\ right-angled\ triangles,\ then\ line\ these\ 2\ and\ the\$
$original\ one\ up, then\ write\ the\ side\ ratios\ in\ order\ to\ get\ x^2\ and\ y^2.$

$The\ 2nd\ way\ is\ to\ draw\ a\ square,\ draw\ 4\ interlocking\ circles$
$whose\ diameters\ are\ the\ sides\ of\ the\ square$
$and\ whose\ centres\ are\ the\ midpoints\ of\ the\ sides.$
$Inside\ the\ square\ are\ now\ 4\ semicircles.$
$You\ can\ now\ draw\ lines\ such\ that\ you\ end\ up\ with$
$4\ identical\ right-angled\ triangles\ and\ a\ central\ square.$
$the\ hypotenuse\ of\ the\ triangles\ are\ the\ sides\ of\ the\ main\ square.$
$You\ can\ then\ quickly\ write\ the\ square\ area\ in\ terms\ of$
$the\ 4\ triangles\ and\ smaller\ central\ square.$

$The\ formula\ is\ the\ expression\ of\ the\ geometry.$

$Plato\ nicely\ summarised\ the\ angle\ calculation.$