QUICK QUESTION!! VCAA 2006 Exam 2

[jane1234]

Okay, just did exam 2 VCAA 2006...

Question 5 a ii

Write down the complex equation of the straight line which passes through the points z1 and -z1 in terms of conj(z1).

So this line has cartesian equation y=x

And conj(z1) = cis(-pi/4)

Answer is |z- conj(z1)| = |z + conj(z1)|

I have no idea how they derived that equation from the cartesian form other than trial and error. Someone please help ASAP!

[jane1234]

BUMP.

Sorry, could someone please explain this as the exam is tomorrow??

[dc302]

If you give me a bit I will have a look, or if someone else is quicker then all the better

[dc302]

Ok, are you familiar with questions that ask you to find the cartesian equation of things like:

|z-1| = |z-i|.... etc? Basically, if you have an expression in this form: |z-a| = |z-b|, where a and b are complex, then it represents a straight line that perpendicularly bisects the points a and b. You can see this because it is saying that, the distance from z to a is the same as the distance from z to b.

So we know that the straight line y=x, is the perpendicular bisector of the two points *z1 and -*z1, so we simply put this into our known expression to get the answer.

[jane1234]

Ok, are you familiar with questions that ask you to find the cartesian equation of things like:

|z-1| = |z-i|.... etc? Basically, if you have an expression in this form: |z-a| = |z-b|, where a and b are complex, then it represents a straight line that perpendicularly bisects the points a and b. You can see this because it is saying that, the distance from z to a is the same as the distance from z to b.

So we know that the straight line y=x, is the perpendicular bisector of the two points *z1 and -*z1, so we simply put this into our known expression to get the answer.

Yes I get it now... Thanks heaps!!

[99.96]

even though dc302 already explained it here's my take on it.

so the line y=x is basically a perpendicular bisector of 2 points in the 4th and 2nd quadrant that and reflections of each other in the y and x axis.
the point conj(z1) is in 4th quad. -conj(z1) is in 2nd quad.
So
its simply
|z-conj(z1)|=|z+conj(z1)|


Just wondering, would this be accepted as an answer aswell:
Im(z)= Re(conj(z1)) x Re(z)

[dc302]

Just wondering, would this be accepted as an answer aswell:
Im(z)= Re(conj(z1)) x Re(z)

I'm not sure what they would accept tbh, but it seems unlikely.