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Math Help - Question(s) about complex numbers.

  1. #1
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    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}.
    Last edited by Captcha; January 1st 2010 at 12:39 PM.
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  2. #2
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    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.
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  3. #3
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    Here is the way to find the principal argument.
    \arg (x + yi) = \left\{ {\begin{array}{rl}<br />
   {\arctan (y/x)} & {,0 < x}  \\<br />
   {\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0}  \\<br />
   {\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0}  \\<br /> <br />
 \end{array} } \right.
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    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.
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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.
    \arg (x + yi) = \left\{ {\begin{array}{rl}<br />
   {\arctan (y/x)} & {,0 < x}  \\<br />
   {\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0}  \\<br />
   {\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0}  \\<br /> <br />
 \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.
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  6. #6
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    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.
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