Question

Let a complex number z=x+ iy satisfies equation |z|4−16|z|2−3z2−3zˉ2+9=0. If a and b are the maximum and minimum value of |z| then ab is equal to _______.

Answer 1

3

Answer 2

We start with the complex number written in polar form:

$$z=r e^{i\theta} \quad \text{with } r=|z|$$

Then

$$z^2=r^2e^{2i\theta}\quad \text{and}\quad \bar{z}^2=r^2e^{-2i\theta}.$$

Given the equation

$$|z|^4-16|z|^2-3z^2-3\bar{z}^2+9=0,$$

substitute $z$ in polar form:

$$r^4-16r^2-3r^2\left(e^{2i\theta}+e^{-2i\theta}\right)+9=0.$$

Since

$$e^{2i\theta}+e^{-2i\theta}=2\cos2\theta,$$

the equation becomes

$$r^4-16r^2-6r^2\cos2\theta+9=0.$$

Rearrange:

$$r^4 - r^2(16+6\cos2\theta) +9=0.$$

For any real $z$, there must exist some $\theta$ so that this equation holds. We isolate $\cos2\theta$:

$$\cos2\theta=\frac{r^4-16r^2+9}{6r^2}.$$

For a solution to exist, we require

$$\left|\frac{r^4-16r^2+9}{6r^2}\right|\le1.$$

This leads to two inequalities:

$$\frac{r^4-16r^2+9}{6r^2}\le1 \quad \Rightarrow \quad r^4-16r^2+9\le6r^2 \quad\Rightarrow\quad r^4-22r^2+9\le0.$$

Let $u=r^2$; then:

$$u^2-22u+9\le0.$$

The quadratic equation $u^2-22u+9=0$ has roots:

$$u=11\pm4\sqrt{7}.$$

Thus,

$$11-4\sqrt{7}\le r^2\le11+4\sqrt{7}.$$

$$\frac{r^4-16r^2+9}{6r^2}\ge-1 \quad \Rightarrow \quad r^4-16r^2+9\ge -6r^2 \quad\Rightarrow\quad r^4-10r^2+9\ge0.$$

Again, with $u=r^2$:

$$u^2-10u+9\ge0.$$

The roots of $u^2-10u+9=0$ are:

$$u=1 \quad \text{and} \quad u=9.$$

Thus,

$$u\le1 \quad \text{or} \quad u\ge9,\quad \text{i.e., } r^2\le1 \text{ or } r^2\ge9.$$

For both conditions to hold, $r^2$ must lie in

$$[11-4\sqrt{7},\,1] \quad \text{or}\quad [9,\,11+4\sqrt{7}].$$

Since $11-4\sqrt{7}\approx 0.417$ and $11+4\sqrt{7}\approx 21.583$, the available values of $r$ form two intervals:

$$r\in\left[\sqrt{11-4\sqrt{7}},\,1\right] \quad \text{and} \quad r\in\left[3,\,\sqrt{11+4\sqrt{7}}\right],$$

where $1=\sqrt{1}$ and $3=\sqrt{9}$.

Thus, the minimum possible $|z|$ is

$$b=\sqrt{11-4\sqrt{7}}$$

and the maximum is

$$a=\sqrt{11+4\sqrt{7}}.$$

Their product is:

$$ab=\sqrt{(11+4\sqrt{7})(11-4\sqrt{7})}=\sqrt{121-16\cdot7}=\sqrt{121-112}=\sqrt{9}=3.$$