Complex Number Locus: Find Locus of z

[unique_pavadrin]

Homework Statement

Find the locus of the point z satisfying:

\left| {\frac{{z - 1 - 2{\bf{i}}}}{{z + 1 + 4{\bf{i}}}}} \right| = 1

2. The attempt at a solution

\begin{array}{l}<br /> \left| {\frac{{z - 1 - 2{\bf{i}}}}{{z + 1 + 4{\bf{i}}}}} \right| = 1 \\ <br /> \left| {\frac{{\left( {x - 1} \right) + \left( {y - 2} \right){\bf{i}}}}{{\left( {x + 1} \right) + \left( {y + 4} \right){\bf{i}}}} \times \frac{{\left( {x + 1} \right) - \left( {y + 4} \right){\bf{i}}}}{{\left( {x + 1} \right) - \left( {y - 4} \right){\bf{i}}}}} \right| = 1 \\ <br /> \left| {\frac{{x^2 + y^2 - 9 + \left( {2y + 2 - 6x} \right){\bf{i}}}}{{\left( {x + 1} \right)^2 + \left( {y + 4} \right)^2 }}} \right| = 1 \\ <br /> \end{array}

im not sure if that is correct but it seemed logical to me at the time, what do i do with the imaginary part of equation?

many thanks,
unique_pavdrin

 

[jhicks]

In line 2 of your attempt at a solution, you switched signs on the y+4 part (to y-4), but it ultimately does nothing to your solution since you didn't follow what you wrote. According to your method, the next step is to apply the definition of magnitude. Have you done this?

 

[unique_pavadrin]

no i am unsure on how to apply the magnitude. do i just square it and then square root it, taking the +ve root?

 

[javierR]

|f(z)|=\sqrt{f f*} where f(z) is a complex function and * denotes complex conjugation.

 

[HallsofIvy]

Homework Statement

Find the locus of the point z satisfying:

\left| {\frac{{z - 1 - 2{\bf{i}}}}{{z + 1 + 4{\bf{i}}}}} \right| = 1

2. The attempt at a solution

\begin{array}{l}<br /> \left| {\frac{{z - 1 - 2{\bf{i}}}}{{z + 1 + 4{\bf{i}}}}} \right| = 1 \\ <br /> \left| {\frac{{\left( {x - 1} \right) + \left( {y - 2} \right){\bf{i}}}}{{\left( {x + 1} \right) + \left( {y + 4} \right){\bf{i}}}} \times \frac{{\left( {x + 1} \right) - \left( {y + 4} \right){\bf{i}}}}{{\left( {x + 1} \right) - \left( {y - 4} \right){\bf{i}}}}} \right| = 1 \\ <br /> \left| {\frac{{x^2 + y^2 - 9 + \left( {2y + 2 - 6x} \right){\bf{i}}}}{{\left( {x + 1} \right)^2 + \left( {y + 4} \right)^2 }}} \right| = 1 \\ <br /> \end{array}
im not sure if that is correct but it seemed logical to me at the time, what do i do with the imaginary part of equation?

many thanks,
unique_pavdrin

I wouldn't do all of that calculation. From
\left| {\frac{{z - 1 - 2{\bf{i}}}}{{z + 1 + 4{\bf{i}}}}} \right| = 1
you have immediately |z-1-2i|= |z+ 1+ 4i| or |z-(1+2i)|= |z-(-1-4i)|.

We can interpret the left side as the distance from z to 1+ 2i and the right side as the distance from z to -1-4i. In other words, the locus is the set of points equidistant from (1,2) and (-1, -4). That is the perpendicular bisector of the line segment between the points.