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Thread: Question(s) about complex numbers.

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

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



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

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

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



    PS. For the last part (as noticed by Archie Meade), I meant: $\displaystyle \arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4}$.
    Last edited by Captcha; Jan 1st 2010 at 12:39 PM.
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  2. #2
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    $\displaystyle There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$

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

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

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

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

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

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

    $\displaystyle You\ must\ make\ this\ evaluation\ first,\ then\ calculate$
    $\displaystyle the\ acute\ angle\ z\ makes\ with\ either\ the\ real\ or\ imaginary\ axis.$
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  3. #3
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    Here is the way to find the principal argument.
    $\displaystyle \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.$
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    $\displaystyle There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$
    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?
    $\displaystyle The\ angle\ is\ taken\ as\ 0\ to\ \pi\ radians\ if\ z\ is\ on\ or\ above\ the\ horizontal\ axis.$
    $\displaystyle It's\ taken\ as\ 0\ to\ -\pi\ for\ z\ below\ the\ horizontal\ axis.$
    $\displaystyle It's\ the\ convention.$$\displaystyle 0\ to\ -\pi\ are\ the\ angles\ from\ 360\ to\ 180\ degrees.$
    Thank you. That is what I have been missing.
    $\displaystyle Arctan\ doesn't\ distinguish\ between\ opposite\ ends\ of\ the\ same\ line,$
    $\displaystyle so\ you\ should\ ask\ yourself\ why\ you\ are\ adding\ 90\ degrees.$
    $\displaystyle Adding\ \frac{\pi}{2}\ radians\ rotates\ a\ line\ through\ 90\ degrees.$
    That was a mistake. What I meant was $\displaystyle \arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4}$
    $\displaystyle From\ the\ real\ and\ imaginary\ parts\ of\ the\ complex\ number,$
    $\displaystyle you\ can\ identify\ it\ as\ being\ between\ 180\ and\ 270\ degrees.$
    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).

    $\displaystyle You\ must\ make\ this\ evaluation\ first,\ then\ calculate$$\displaystyle 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.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Here is the way to find the principal argument.
    $\displaystyle \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 $\displaystyle -1-i$, $\displaystyle \arg(z) = \arctan\left(\frac{-1}{-1}\right)-\pi = \arctan\left(1\right)-\pi = \frac{\pi}{4}-\pi = \frac{\pi-4\pi}{4} = -\frac{3\pi}{4}$. Brilliant. Thanks.
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  6. #6
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    $\displaystyle Hi\ Captcha,$

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

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

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

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

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