About the roots of the equation z2 + bz + c = 0

1. Where are the roots of the equation z2 + bz + c = 0 ?

Play with the figure: push buttons and move z in the plane.
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Problems 1

Reset by using the computer key 'R' (when the figure is clicked to be active).
To choose the equation coefficients b and c at random, move the four yellow points on the horizontal bottom lines.
a) What is the value of b?     Answer a): b = (      ) + i(      )
b) What is the value of c?     Answer b): c = (      ) + i(      )
c) Click the button Show z^2 + bz + c.
Find the two solutions z1 and z2 of the equation
z2 + bz + c = 0.
Hint. Move the free variable red point z on the plane area.
Answer c1): z1 = (      ) + i(      )
Answer c2): z2 = (      ) + i(      )
d) Compute z1 + z2 and z1 z2.
Answer d1): z1 + z2 = (      ) + i(      )
Answer d2): z1 z2 = (      ) + i(      )

2. About the roots and coefficients of z2 + bz + c = 0

Play with the figure: push buttons and move z in the plane.
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Problems 2

Reset by using the computer key 'R' (when the figure is clicked to be active).
Choose again b and c at random.
2a) What is the value of b?     Answer 2a): b = (      ) + i(      )
2b) What is the value of c?     Answer 2b): c = (      ) + i(      )
2c) Click the buttons Show z^2 and Show - bz - c.
Find the two solutions z1 and z2 of the equation
z2 + bz + c = 0.
Answer 2c1): z1 = (      ) + i(      )
Answer 2c2): z2 = (      ) + i(      )
2d) Compute z1 + z2 and z1 z2.
Answer 2d1): z1 + z2 = (      ) + i(      )
Answer 2d2): z1 z2 = (      ) + i(      )
How can you explain the relations between z1 + z2, z1 z2, b and c?
Answer:

3. The loci of f(z) : = z2 + bz + c on segments

Introduction to tracing

In the dynamic figure beside you see a variable complex number z and its image f(z) in the complex plane. Move z. When you click Show z^2 + bz + c and move z, you seen traces that the image point f(z) leaves behind. Tracing is stopped by the Hide button, but cleared from the screen by pushing the red cross on the lower right corner.
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Problems 3

3a) Try to see by tracing the image of the real number set.
Hint. Use tracing and move z along the horizontal axis, the real part x =  Re z axis.
Finally, see the image of the x-axis under f by clicking the button Image of x-axis.
3b) Reset by using the computer key 'R', and choose again b and c at random (or leave them as they are).
Click the buttons Show z1 and z2 and Show segment z1z2. You should see the segment connecting the zeros of f, i.e. the two numbers z1 and z2, for which f(z1) = 0 = f(z2). Try with z. You also see a movable red point on the segment; it's there to help you.
Click the green buttons Segment and Animate z on segment (the animation is stopped by the same button).
When z describes the segment, the same time f(z) = z2 + bz + c is describing its image, the curve called locus.
Click on the green Locus button to see the image { f(z) | z on the segment }.
Change the segment using its endpoints A and B. The locus is nearly always an arc of a parabola.
3c) What is the shape of the locus when it is not an arc of parabola?
Answer 3c):
3d) When does it happen (use the red point on the segment)?
Answer 3d):
3e) Click on Show (z1+z2)/2. Can you see whether you were right?
Answer 3e):
3f) Hide z1, z2 and [1/2](z1 + z2). Can you imagine a new method of finding z1 and z2, using the segment and its image under f?
Answer:


3g) Difficult question, not compulsory:
Prove that your answers above are always true.
Answer:

4. The loci of f(z) : = z2 + bz + c on circles

Play with the figure: push buttons and move z in the plane.
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Problems 4

4a) Try to see by tracing the image of the unit circle (UC).
See the locus of the unit circle under f by clicking the button Image of UC.
4b) Reset by using the computer key 'R', and choose b and c at random (or leave them as they are).
Click the buttons Show segment z1z2 and Show (z1+z2)/2.
How can you verify that the numbers z1 and z2 really are the zeros of f(z) : = z2 + bz + c?
Answer 4b):
4c). Click the gray buttons Show circle and Animate z on circle.
When z describes the circle, the same time f(z) describes its image, the curve called locus. Try tracing, stop it and clear the trace.
Click on the gray Locus button, and change the center and radius of the circle.
Are the loci of circles always circles?
Answer 4c):
4d) What happens when you place the center of circle of animation on top of the midpoint of the segment connecting the zeros of f (the point [1/2](z1 + z2)) ?
Answer 4d):
4e) Place the circle of animation on top of the zeros of f. Can you imagine how to find the zeros just by using the loci of circles?
Answer 4e):

4f) The locus has a double point if there are two distinct points p1 and p2 so that f(p1) = f(p2). Find out the circles whose locus has double points.
Answer 4f):



4g) Difficult question, not compulsory:
Prove that the locus of a circle under a polynomial f(z) = z2 + bz + c has a double point if and only if the midpoint [1/2](z1 + z2) of the zeros of f is inside the circle.
Answer 4g):