Question

Complex numbers help. *question attached below*

Answer 1

So I have already found the Cartesian equation of the locus of z, which I got as: 3x^2 - 3y^2 +10x +3 =0. But, I am having some difficulties to the Argand diagram and shading bit. Can I have some help, please?

Answer 2

Slight mistake:\[\begin{align*}
\left|\frac{z-1}{z+1}\right|&=\frac{\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}}&\text{where }z=x+iy\\\\
2\sqrt{(x+1)^2+y^2}&=\sqrt{(x-1)^2+y^2}\\\\
4(x+1)^2+4y^2&=(x-1)^2+y^2\\\\
4x^2+8x+4+4y^2&=x^2-2x+1+y^2\\\\
3x^2+10x+3\color{red}{+}3y^2&=0\end{align*}\]
To see what this equation represents, see what happens when you complete the square(s):
\[\begin{align*}3x^2+10x+3+3y^2&=0\\\\
3\left(x^2+\frac{10}{3}x\right)+3(y-0)^2&=-3\\\\
3\left(x^2+\frac{10}{3}x+\frac{25}{9}-\frac{25}{9}\right)+3(y-0)^2&=-3\\\\
3\left(\left(x+\frac{5}{3}\right)^2-\frac{25}{9}\right)+3(y-0)^2&=-3\\\\
3\left(x+\frac{5}{3}\right)^2-\frac{25}{3}+3(y-0)^2&=-3\\\\
3\left(x+\frac{5}{3}\right)^2+3(y-0)^2&=\frac{16}{3}\\\\
\left(x+\frac{5}{3}\right)^2+(y-0)^2&=\frac{16}{9}
\end{align*}\]
which is the equation of a circle of radius \(\dfrac{4}{3}\) centered at \(\left(-\dfrac{5}{3},0\right)\).

Answer 3

Oh, my bad. Thank you for the correction :) @SithsAndGiggles

Answer 4

How do I go about sketching this locus in an Argand diagram?

Answer 5

An Argand diagram is hardly different from a regular x-y plane. The horizontal axis represents the real part of a complex number, and the vertical axis represents the imaginary part.

If you can plot a circle given its center and radius, you're halfway there.

As for the region you have to shade: Pick any \(z\) in the plane. For instance, suppose \(z=1\), or in terms of a coordinate point, \((1,0)\). Does this point satisfy the given inequality?
\[\left|\frac{1-1}{1+1}\right|=0<2\]
The answer is no. Check any other point outside the circle, and you'll get a similar result. This means you will be shading the interior of the circle, as any point inside satisfies \(\left|\dfrac{z-1}{z+1}\right|>2\) (except of course \(z=-1\), i.e. \((-1,0)\)). As an example, suppose \(z=-\dfrac{5}{3}\) (the center), then
\[\left|\frac{-\frac{5}{3}-1}{-\frac{5}{3}+1}\right|=4>2\]
Do you consider values on the boundary of the circle? Let \(z=-\dfrac{1}{3}\), which lies on the boundary directly to the right of the center.
\[\left|\frac{-\frac{1}{3}-1}{-\frac{1}{3}+1}\right|=2\not>2\]
You do not.

Answer 6

You're not quite done yet, because you have to account for the other condition that \(0<\arg z<\dfrac{3\pi}{4}\).

Any number in this region will satisfy this:

Find the intersection of this region with the disk.