Complex line equation with condition |a| = |b|

[PcumP_Ravenclaw]

Dear All,
Please help me clear some doubts about Theorem 3.3.1 in the 1st attachment.
The condition $|a| = |b|$ has only 8 cases right? ${ x+iy. x - iy, -x + iy, -x - iy, y + ix, y - ix, -y + ix, -y - ix }$

so for the condition $|a| = |b|$ and $b \bar c = \bar a c$ in (2) and (3) in the attachement, what must $b$ and $\bar a$ be for them to to satisfy this equation given that C is a complex number of the form $Cx + iCy$.

why is there a line of solutions in (3)? usually you only get one imaginary and one real value for z right??

In the 2nd attachement, I have tried to do question 3. what does in the direction of b mean? does it pass through b also?? c =0 right? can you please give examples of the complex number b?

Danke...

 

[Mark44]

Dear All,
Please help me clear some doubts about Theorem 3.3.1 in the 1st attachment.
The condition $|a| = |b|$ has only 8 cases right? ${ x+iy. x - iy, -x + iy, -x - iy, y + ix, y - ix, -y + ix, -y - ix }$

No, and how are x + iy, x - iy, etc. conditions? These are complex numbers. I don't see that they are conditions in any way. For example, consider a = 0 + 1i and b = -1/2 + (√3/2)i. These two complex numbers have the same magnitude, but I don't see how they are included in your eight cases, whatever it is you mean by them.

so for the condition $|a| = |b|$ and $b \bar c = \bar a c$ in (2) and (3) in the attachement, what must $b$ and $\bar a$ be for them to to satisfy this equation given that C is a complex number of the form $Cx + iCy$.

why is there a line of solutions in (3)? usually you only get one imaginary and one real value for z right??

In the 2nd attachement, I have tried to do question 3. what does in the direction of b mean?

b is a complex number, so you can think of it as being a vector, with its tail at the origin and its head at the point in the complex point (b₁, b₂). What direction does it point?

does it pass through b also?? c =0 right? can you please give examples of the complex number b?

Danke...