Complex Numbers and Geometry (1 Viewer)

CM_Tutor

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Here are a few more questions for people to ponder. I know that each can be done algebraically, and this is better than not being able to solve them at all, but all are more easily done geometrically.

1. P and Q are points on the Argand Diagram representing the complex numbers z and w, respectively, and z / w is purely imaginary. Draw a diagram to represent this information, and use it to prove that |z + w| = |z - w|.

2. P and Q are points on the Argand Diagram representing the complex numbers z and w, respectively, and
|z| = |w|. Draw a diagram to represent this information, and use it to prove that (z + w) / (z - w) is purely imaginary.

3. z<sub>0</sub> is a complex number represented by the point P on the Argand Diagram, and Q represents the point iz<sub>0</sub>.
(a) Draw a sketch of the locus of z if |z - z<sub>0</sub>| = |z - iz<sub>0</sub>|. Include the points P and Q on your sketch.
(b) Draw a sketch of the locus of z if arg(z - z<sub>0</sub>) = arg(iz<sub>0</sub>). Include the points P and Q on your sketch.
(c) The loci in (a) and (b) meet at a point R. Find the complex number represented by the point R.

4. If z is any complex number satisfying |z| = 1, show that 1 <= |z + 2| <= 3, and |arg(z + 2)| <= pi / 6.

5. If z<sub>0</sub> is a fixed complex number and R is a positive constant, describe the locus of z if
z * z(bar) + z * z<sub>0</sub>(bar) + z(bar) * z<sub>0</sub> + z<sub>0</sub> * z<sub>0</sub>(bar) = R<sup>2</sup>
 

CM_Tutor

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27 views and not a single attempt yet. Am I asking something that's too hard, or do people not want the extra practice questions, or ...?
 

Calculon

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Heres 3 a

z remains on the x-axis, hence its principal argument is 0 or pi and it is purely real
 
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Calculon

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Here 3b

z stays on the y-axis hence its principle argument is pi/2 or -pi/2 and it is purely imaginary
 
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Giant Lobster

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1) rectangle >> diagonals are equal
2) rhombus >> diagonals perpendicular
3) ugh... this needs pen and paper :(
4) circle radius 1 centre origin. from -2,0 min distance is 1, max is 3. hence first part. 2nd part, the tangents to circle from -2,0 make angle with positive x axis varying btn pi/6 and -pi/6 (you left that one out i think)
5) |z+z_0| = R. Circle centre -z_0, radius R.
 

CM_Tutor

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Sorry, Calculon, 3a and 3b are wrong.
Originally posted by Giant Lobster
4) circle radius 1 centre origin. from -2,0 min distance is 1, max is 3. hence first part. 2nd part, the tangents to circle from -2,0 make angle with positive x axis varying btn pi/6 and -pi/6 (you left that one out i think)
No, I didn't. The question said |arg(z + 2)| <= pi / 6, not arg(z + 2) <= pi / 6.
 
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Calculon

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^What he said^
I'll have another go when I'm not so tired
 

Faera

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3a.

line = perpendicular to the line joining P and Q, going through (0,0)

3b.

Line parallel to vector OQ going through P.

?
 

CM_Tutor

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Originally posted by Faera
3a.

line = perpendicular to the line joining P and Q, going through (0,0)
and more importantly, through the midpoint of PQ
3b.

Line parallel to vector OQ going through P.

?
Almost. actually it's a ray parallel to vector OQ, starting at (but not including) P.

For (c), you'll need the diagram and some geometry, or some neat complex number algebra.
 

Faera

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3c.

Let the lines/rays from q's 3a,b, intersect at point R.
RO perpendicular PQ (from 3a)
Since RO, PQ are the diagonals of QRPO, the QRPO must be a rhombus (since QO and PO are equal, it cannot be a kite).
Therefore, PR = OQ
and R = vector OP + vector OQ.
that is, R = z(0) + i(z(0))
R = z(0) * ( 1 + i )

(i dont know how to do that subscript thing... -.-')
 

CM_Tutor

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Subscript is < sub > whatever < /sub >, but without the spaces

Your answer for R is correct - the algebra way (for anyone whose interested) means solving simulataneously the two equations |z - z<sub>0</sub>| = |z - iz<sub>0</sub>| and arg(z - z<sub>0</sub>) = arg(iz<sub>0</sub>). This should be done without putting z = x + iy.
 

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