Question

The number of solution(s) of the equation $z^2=4z+|z|^2+$ $\frac{16}{|z|^3}$ is (where z=x+iy, x, y ∈ R, $i^2$ = -1 and x ≠ 2)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

Solution:

Given the equation

$$z^2 = 4z + |z|^2 + \frac{16}{|z|^3}$$

with $z = re^{i\theta}$ (where $r = |z| > 0$ and $x \neq 2$), rewrite the equation as:

$$r^2e^{2i\theta} = 4re^{i\theta} + r^2 + \frac{16}{r^3}.$$

Dividing through by $r$ (since $r>0$) we obtain

$$r e^{2i\theta} = 4e^{i\theta} + r + \frac{16}{r^4}.$$

Step 1. Equate the imaginary parts:

Write

$$e^{2i\theta} = \cos 2\theta + i\sin 2\theta,\quad e^{i\theta} = \cos\theta + i\sin\theta.$$

Then the imaginary part gives:

$$r\sin 2\theta = 4\sin\theta.$$

Since $\sin2\theta = 2\sin\theta\cos\theta$, we have

$$2r\sin\theta\cos\theta = 4\sin\theta.$$

There are two cases:

Case 1: $\sin\theta \neq 0$ Dividing by $\sin\theta$:

$$2r\cos\theta = 4 \quad\Longrightarrow\quad r\cos\theta = 2.$$

Substituting this into the real part (next step) will lead to a contradiction (a negative value for $r^5$ that is impossible for $r>0$). Thus no solution arises from this case.

Case 2: $\sin\theta = 0$ This implies $\theta = 0$ or $\theta = \pi$.

Subcase 2a: $\theta = 0$ (so $z = r$, a positive real number) The equation becomes:

$$r^2 = 4r + r^2 + \frac{16}{r^3}.$$

Cancel $r^2$ from both sides:

$$0 = 4r + \frac{16}{r^3}.$$

Multiplying by $r^3$:

$$0 = 4r^4 + 16 \quad\Longrightarrow\quad 4r^4 = -16,$$

which is impossible for $r>0$.

Subcase 2b: $\theta = \pi$ (so $z = -r$) Now $e^{i\pi} = -1$ and $e^{2i\pi}=1$. The equation becomes:

$$r^2 = 4(-r) + r^2 + \frac{16}{r^3}.$$

Cancelling $r^2$ from both sides:

$$0 = -4r + \frac{16}{r^3}.$$

Multiply by $r^3$:

$$0 = -4r^4 + 16 \quad\Longrightarrow\quad 4r^4 = 16 \quad\Longrightarrow\quad r^4 = 4.$$

Hence,

$$r = \sqrt[4]{4} = 4^{1/4} = \sqrt{2}.$$

This gives the unique solution:

$$z = -\sqrt{2}.$$

Conclusion: There is exactly 1 solution.