Question

The number of solution(s) of the equation $z^2=4z+|z|^2$

$\frac{16}{|z|^3}$ is (where $z=x+iy, x, y \in R, i^2=-1$ and $x \neq 2$)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

Solution:

We rewrite the given equation as

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

where $z = re^{i\theta}$ and $r = |z| > 0$.

Expressing in polar form:

$$z^2 = r^2 e^{2i\theta} \quad \text{and} \quad 4z = 4r e^{i\theta}.$$

Thus the equation becomes

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

Equate the imaginary parts:

$$r^2 \sin2\theta = 4r \sin\theta.$$

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

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

If $\sin\theta \neq 0$, divide by $\sin\theta$ (since $\sin\theta$ is nonzero):

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

Substituting this into the real part (where $\cos2\theta = 2\cos^2\theta-1$) eventually leads to a contradiction (since it would require $r^3$ to be negative for positive $r$). Hence, we must have

$$\sin\theta = 0.$$

Thus, $\theta = 0$ or $\theta = \pi$.

Case 1: $\theta = 0$
Here, $z = r$ (a positive real). The equation becomes:

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

Multiplying by $r$ (with $r>0$):

$$r^3 = 4r^2 + 16 \quad \Longrightarrow \quad r^3 - 4r^2 - 16 = 0.$$

This cubic in $r$ has exactly one positive root (by the Intermediate Value Theorem). Also, $x = r \neq 2$ is maintained.

Case 2: $\theta = \pi$
Here, $z = -r$ (with $r>0$). Now:

$$r^2 = -4r + \frac{16}{r},$$

Multiplying by $r$:

$$r^3 = -4r^2 + 16 \quad \Longrightarrow \quad r^3 + 4r^2 - 16 = 0.$$

This equation also has exactly one positive root.

Conclusion:
There are exactly 2 solutions corresponding to $\theta = 0$ and $\theta = \pi$.

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Minimal Explanation of the Solution:
Let $z = re^{i\theta}$. Equate imaginary parts; $\sin\theta$ must be 0, so $\theta = 0$ or $\pi$. For each, the resulting real equation in $r$ has one positive solution. Hence, the original equation has 2 solutions.