I am attempting to determine the locus of z for arg &#x2061;<!-- ⁡ --> ( z 2

junoonib89p4 2022-05-07 Answered
I am attempting to determine the locus of z for arg ( z 2 + 1 ). I do know that since z 2 + 1 = ( z + i ) ( z i ), I could write arg ( z 2 + 1 ) = arg ( z + i ) + arg ( z i ) but I don’t really know how to figure out the locus of the points from here. How should I approach this? Is there a relationship with the difference in two arguments (arc of a circle)? Thank you!
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Answers (1)

Gary Salinas
Answered 2022-05-08 Author has 7 answers
Let z = x + i y with x , y R , and let arg ( z 2 + 1 ) = α .

1. If α = ± π / 2 , then z 2 + 1 must lie on the imaginary axis, so z 2 + 1 = i b for some b R , which in cartesian coodinates translates to x 2 y 2 + 2 i x y + 1 = i b , so the locus is part of the hyperbola x 2 y 2 + 1 = 0 .
2. If α ± π / 2 , then a = tan ( α ) = Im ( z ) / Re ( z ) is well defined, and the equation becomes:
2 x y x 2 y 2 + 1 = a a x 2 2 x y a y 2 + a = 0
This is a quadratic equation with discriminant 2 2 4 a ( a ) > 0 , so the locus is again part of a hyperbola.

In both cases, the actual locus is the part of the hyperbola lying in the quadrant determined by α .
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I am aware that it must be part of an arc of a circle that passes through the points a and b. The argument means that the vector ( z a ) leads the vector ( z b ) by θ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to
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