Modulus and Argument (FP1) Watch

AvM4N
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What does the loci, for example, lzl= ( z-6 ) look like.
Also, how do you find the the point of intersection of two loci- e.g.- find the complex number represented by the intersection of c1 (l z+2 l=2) and c2 (arg(z+2)= 5/6pi
I've drawn them both out and the answer is -(2+ root3) +i
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Zacken
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(Original post by AvM4N)
What does the loci, for example, lzl= ( z-6 ) look like.
Also, how do you find the the point of intersection of two loci- e.g.- find the complex number represented by the intersection of c1 (l z+2 l=2) and c2 (arg(z+2)= 5/6pi
I've drawn them both out and the answer is -(2+ root3) +i
Do you mean |z| = |z-6|? If so, that's the perpendicular bisector of the point z=0 and z=6.

Oh, and to answer your next question, draw the diagrams and look for a geometrical solution. If you can't find one, then use cartesian forms:

|z+2| = 2 is a circle centre (0, -2) and radius 2, so: x + (y+2)^2 = 4.

And arg(z+2) = 5pi/6 is a half line with cartesian equation \frac{y}{x+2} = -\frac{\sqrt{3}}{3}, re-arrange for y and plug it into he above equation to get a quadratic in x than you can then solve.
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AvM4N
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(Original post by Zacken)
Do you mean |z| = |z-6|? If so, that's the perpendicular bisector of the point z=0 and z=6.

Oh, and to answer your next question, draw the diagrams and look for a geometrical solution. If you can't find one, then use cartesian forms:

|z+2| = 2 is a circle centre (0, -2) and radius 2, so: x + (y+2)^2 = 4.

And arg(z+2) = 5pi/6 is a half line with cartesian equation \frac{y}{x+2} = -\frac{\sqrt{3}}{3}, re-arrange for y and plug it into he above equation to get a quadratic in x than you can then solve.
ahh, thanks for your help
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Zacken
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(Original post by AvM4N)
ahh, thanks for your help
No worries.
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