complex locus (1 Viewer)

mabc1

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given Arg(z-1) = 2Arg(z), deduce locus of z is a circle
part ii. find z in mod arg form if z also satisfies |z| = |z-1|
 

fan96

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Since , the equation can be rearranged to



Note that an argument of zero means that could be a positive real number , as and any real that is less than 1 produces a negative number with argument .

This means there are restrictions on the real and negative parts of .

Also, the identity might be useful.

The locus is not a circle, because as stated before any satisfies the equation.
 
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mrbunton

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for i) z-1/z^2 = a+0i, because arg is 0, therefore a number real. equate the imaginary part of the z-1/z^2 to 0 and get it into the form of a circle.
for ii) i think it creates a straight line; so simultaneously solve the line and the circle to get the intended values.
im too lazy to do it thou but thats how to solve it.
 
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altSwift

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Since , the equation can be rearranged to



Note that an argument of zero means that could be a positive real number , as and any real that is less than 1 produces a negative number with argument .

This means there are restrictions on the real and negative parts of .

Also, the identity might be useful.

The locus is not a circle, because as stated before any satisfies the equation.
Can you pleaes explain how you rearranged it to get
 

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