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I have two regions in the complex plane, defined by $|z-1|=1$ and $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$, $\mathrm{Re}(z)>0$.

I am being asked to find and sketch the image of those regions under the mapping $f(z) = \sqrt{z}$ .

My attempt

We have $f(z) = \sqrt{r}e^{i\frac{\theta}{2}}$ and $|z-1|=1 \iff r = 2\cos(\theta)$

If $|z-1|=1 $, then: $$f(z) = \sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right) + i\sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right)$$

It means that $f(z)$ lies on the trace of the curve $\left(\sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right); \sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right) \right)$ if $|z-1| = 1.$

Am I right so far?

I couldn't recognize this plane curve.

And I don't know how to proceed in the region $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$. How can I do it?

Michael McGovern
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Santos
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  • The function $\sqrt{z}$ is undefined at $z=0$, how to map the circle that passes through the origin ? – Zaid Alyafeai Apr 19 '17 at 16:30
  • You have also to define a branch of the square root since it is multi-valued. The principal root of a hyperbola doesn't make sense because it passes through the branch cut. – Zaid Alyafeai Apr 19 '17 at 16:36

2 Answers2

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  1. $|z-1|=1$ is the circle with center $(1,0)$, thus $z=(\cos\theta+1)+i\sin\theta$. In polar coordinates, you have $z=\big(\sqrt{(1+\cos\theta)^2+\sin\theta^2}\big)e^{i\theta}=\big(\sqrt{2+2\cos\theta}\big)e^{i\theta}$. Thus:

$$f(z)=\big( \root4\of{2+2\cos\theta} \big)e^{\frac{i\theta}{2}}$$

A numerical approach showed it should look like a flower with 4 petals.

  1. $(\Im(z))^2=(\Re(z))^2-1$ in polar coordinates yields:

$$r^2\sin^2\theta=r^2\cos^2\theta-1\implies r^2\cos2\theta=1$$

Thus $r=\sqrt{1-\cos2\theta}$, so $z=\big(\sqrt{1-\cos2\theta}\big)e^{i\theta}$. Then, applying $f$:

$$f(z)=\big(\root4\of{(1-\cos2\theta)}\big)e^{\frac{i\theta}{2}}$$

Numerical results shows that it should look like an $8$.

For sketching, calculating a few points might come in handy:

  • For which values of $\theta$ does the graph cross the real axis and the imaginary axis?
  • For which values of $\theta$ does the graph cross the origin?
  • For which values of $\theta$ does it attain maximums and minimums (that is, $\frac{dy}{dx}=0$)?
  • For which values of $\theta$ do we have vertical slope (that is, $\frac{dx}{dy}=0$)?

Plot those points into the plane and draw a line that passes through each point in increasing order of $\theta$.

Note that for the last 2 types of points, $x=\Re(f(z))$ and $y=\Im(f(z))$; so, you can use parametric derivatives with respect to $\theta$.

AspiringMathematician
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Possible suggestion:

Let $z = x + yi$. We will emulate this problem in the Cartesian plane.

By transforming $Re(z)^2 = Im(z)^2 - 1$ under $f$, we are taking the root of each point. Conversely, $z^2 = (x^2 - y^2) + (2xy)i$ satisfies $Re(z^2)^2 = Im(z^2)^2 - 1$ instead for every point $z$ on the transformed curve.

This is the Cartesian graph $x^4 - 2x^2y^2 + y^4 = 4x^2y^2 - 1$

I.E. $x^4 - 6x^2y^2 + y^4 = - 1$

The first problem can be done similarly. The transformed curve is $|z^2 - 1| = 1$ This translates to the Cartesian plane as $4x^2y^2 + x^4 + y^4 + 1 - 2x^2y^2 + 2y^2 - 2x^2 = 1$

I.E. $0 = 2x^2y^2 + x^4 + y^4 + 2y^2 - 2x^2$

TCiur
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