graph of complex number

**BabyMilo** · June 10th 2010, 09:57 AM

I am sorry if i posted in the wrong subforum

here is the question, i know to draw the graph but duno how to do part ii)

any help?

thanks!!

**slider142** · June 10th 2010, 03:00 PM

From your graph, what do you know about the imaginary part of points in C1 ? If the point is also in C2, you have its argument. Use these two pieces of information to find the real part of your point.

**Prove It** · June 10th 2010, 05:06 PM

Originally Posted by BabyMilo

I am sorry if i posted in the wrong subforum

here is the question, i know to draw the graph but duno how to do part ii)

any help?

thanks!!

$|z| = |z - 4i|$

$|x + iy| = |x + i(y - 4)|$

$\sqrt{x^2 + y^2} = \sqrt{x^2 + (y - 4)^2}$

$x^2 + y^2 = x^2 + (y - 4)^2$

$y^2 = (y - 4)^2$

$y^2 = y^2 - 8y + 16$

$0 = -8y + 16$

$8y = 16$

$y = 2$ .

You also know that $\tan{\theta} = \frac{y}{x}$ and $\theta = \frac{\pi}{6}$ .

Putting it together, with $y = 2$ gives

$\tan{\frac{\pi}{6}} = \frac{2}{x}$

$\frac{1}{\sqrt{3}} = \frac{2}{x}$

$\frac{x}{\sqrt{3}} = 2$

$x = 2\sqrt{3}$ .

So therefore, the point of intersection is

$2\sqrt{3} + 2i$ .

**BabyMilo** · June 10th 2010, 09:26 PM

Originally Posted by Prove It

$|z| = |z - 4i|$

$|x + iy| = |x + i(y - 4)|$

$\sqrt{x^2 + y^2} = \sqrt{x^2 + (y - 4)^2}$

$x^2 + y^2 = x^2 + (y - 4)^2$

$y^2 = (y - 4)^2$

$y^2 = y^2 - 8y + 16$

$0 = -8y + 16$

$8y = 16$

$y = 2$ .

You also know that $\tan{\theta} = \frac{y}{x}$ and $\theta = \frac{\pi}{6}$ .

Putting it together, with $y = 2$ gives

$\tan{\frac{\pi}{6}} = \frac{2}{x}$

$\frac{1}{\sqrt{3}} = \frac{2}{x}$

$\frac{x}{\sqrt{3}} = 2$

$x = 2\sqrt{3}$ .

So therefore, the point of intersection is

$2\sqrt{3} + 2i$ .

thanks...have an exam today.
thanks for the help!

**mr fantastic** · June 10th 2010, 10:22 PM

Originally Posted by BabyMilo

I am sorry if i posted in the wrong subforum

here is the question, i know to draw the graph but duno how to do part ii)

any help?

thanks!!

You should know the geometric interpretation of these two loci:

$|z - z_1| = |z - z_2|$ defines the perpendicular bisector of the line segment joining $z_1$ and $z_2$ .

$Arg(z - z_1) = \theta$ defines a ray with terminus $z_1$ (not included in the locus because Arg(0) is not defined) and inclined at an angle $\theta$ to the horizontal.

**slider142** · June 17th 2010, 01:27 PM

Just an additional point of view. For future reference, try to visualize the geometry beforehand using the geometric meaning of the equations. For example, your first equation states that |z - 0| = |z - 4i|. This means that the distance between z and 0 is the same as the distance between z and 4i. This is describing the horizontal line of points with imaginary part 2 (right between 0 and 4i).
The second equation was covered by the previous poster as a ray. You can then find the intersection point by using basic trigonometry, as the ray, horizontal line, and imaginary axis form a right triangle.
While you can work it out algebraically, it is often much faster to see the full geometric picture first, using the concepts of distance and angle in the complex plane as analogues to modulus and argument, respectively. Of course, this only helps if you have a solid grounding in plane geometry as well (ie., you should know the conics as geometric descriptions of loci, not just their Cartesian equations).

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