1
$\begingroup$

Trying to graph $Im((z + 4 - 4i)^2) = 1$ on the complex plane.

I'm having trouble interpreting this equation to sketch it by hand. Fully expanding this is not helpful (or practical) when trying to sketch this by hand.

I understand how to graph complex equations such as $|z|^2 = 25$ (a circle of radius 5) and $|z - 2 - 2i| = 3$ (circle of radius 3 with centre $2 + 2i$). You can substitute $z = x + iy$ for all of these types of problems; but, I generally understand their geometric form due to the meaning of the modulus and so on.

As for this hyperbola; I'm not sure of the approaches / steps required to sketch this by hand.

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

As you mention, we just plug in $z=x+iy$ to get $(z+4-4i)^2 = \{(x+4)+(y-4)i\}^2$.

The imaginary part is $2(x+4)(y-4)$ so we have

$$2(x+4)(y-4)=1$$

Your plot should look like a hyperbola.

$\endgroup$
0
$\begingroup$

enter image description hereThe substiution $z=x+iy$ works just as well in this example. Do the algebra ... we get $2(x+4)(y-4)=1$. So the solution is a rectangular hyperbola with asymptotes at $x=-4$ and $y=4$.

$\endgroup$
2
  • $\begingroup$ $p=(x+i y+4-4 i)^2=x^2+i (2 x y-8 x+8 y-32)+8 x-y^2+8 y\quad \Im(p)=1$ leads to $2 x y-8 x+8 y-32=1$ that is $y=\dfrac{8 x+33}{2 (x+4)}$ $\endgroup$
    – Raffaele
    Commented Aug 23, 2017 at 18:00
  • $\begingroup$ @Raffaele If you require that $y$ is the subject of the formula then it is probably better to write it as \begin{eqnarray*} y=4 +\frac{1}{2(x+4)}. \end{eqnarray*} The question ask for the graph, so it is better to leave it in the form $XY=c^2$ where it can be recognised as a rectangular hyperbola whose asymptotes can be easily identified. $\endgroup$
    – Donald Splutterwit
    Commented Aug 23, 2017 at 19:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .