Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\displaystyle\sum_{z \in S}|z|^2$ is equal to:
A) $\dfrac{5}{64}$
B) $\dfrac{1}{16}$
C) $\dfrac{7}{64}$
D) $\dfrac{3}{16}$ ✓
✓
Solution Steps
1
Write z = x + iy, expand 4z² + z̄ = 0
$4(x+iy)^2 + (x-iy) = 0$
$4(x^2 – y^2 + 2ixy) + x – iy = 0$
Real part: $4x^2 – 4y^2 + x = 0 \;\cdots(1)$
Imaginary part: $8xy – y = 0 \implies y(8x-1)=0 \;\cdots(2)$
2
Case 1: y = 0 (real solutions)
Substitute into (1): $4x^2 + x = 0 \implies x(4x+1)=0$
$x = 0 \Rightarrow z_1 = 0$
$x = -\tfrac{1}{4} \Rightarrow z_2 = -\tfrac{1}{4}$
$x = -\tfrac{1}{4} \Rightarrow z_2 = -\tfrac{1}{4}$
3
Case 2: x = 1/8 (complex solutions)
Substitute $x = \frac{1}{8}$ into (1):
$4\cdot\frac{1}{64} – 4y^2 + \frac{1}{8} = 0 \implies \frac{1}{16} + \frac{1}{8} = 4y^2$
$4y^2 = \frac{3}{16} \implies y^2 = \frac{3}{64} \implies y = \pm\frac{\sqrt{3}}{8}$
$z_3 = \tfrac{1}{8}+\tfrac{\sqrt{3}}{8}i, \quad z_4 = \tfrac{1}{8}-\tfrac{\sqrt{3}}{8}i$
4
Compute |z|² for each element of S
| z | |z|² |
|---|---|
| $0$ | $0$ |
| $-\frac{1}{4}$ | $\frac{1}{16}$ |
| $\frac{1}{8}+\frac{\sqrt{3}}{8}i$ | $\frac{1}{64}+\frac{3}{64}=\frac{4}{64}=\frac{1}{16}$ |
| $\frac{1}{8}-\frac{\sqrt{3}}{8}i$ | $\frac{1}{64}+\frac{3}{64}=\frac{1}{16}$ |
5
Sum up all |z|²
$$\sum_{z\in S}|z|^2 = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$$
✓ Answer: D) 3/16
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Key Insight
Imaginary part → y(8x-1)=0 → 2 cases. Real part → 4 solutions total. Σ|z|²=3/16.
📐
Theory: Complex Numbers — Key Formulas
Essential Identities
| Identity | Formula |
|---|---|
| Modulus | $|z|^2 = x^2+y^2 = z\cdot\bar{z}$ |
| Conjugate | $\bar{z}=x-iy$; $\overline{z^n}=\bar{z}^n$ |
| $z+\bar{z}$ | $2\text{Re}(z)=2x$ |
| $z-\bar{z}$ | $2i\cdot\text{Im}(z)=2iy$ |
| $z^2$ | $(x^2-y^2)+2ixy$ |
| $|z_1 z_2|$ | $|z_1||z_2|$ |
| Triangle inequality | $|z_1+z_2|\leq|z_1|+|z_2|$ |
Strategy: Solving f(z, z̄) = 0
When equation mixes $z$ and $\bar{z}$:
Step 1: Write $z = x+iy$, $\bar{z}=x-iy$
Step 2: Expand fully
Step 3: Separate Real and Imaginary parts = 0
Step 4: Solve the system of equations
Alternative: Write $\bar{z} = |z|^2/z$ (for $z \neq 0$) and solve as pure $z$-equation.
Step 1: Write $z = x+iy$, $\bar{z}=x-iy$
Step 2: Expand fully
Step 3: Separate Real and Imaginary parts = 0
Step 4: Solve the system of equations
Alternative: Write $\bar{z} = |z|^2/z$ (for $z \neq 0$) and solve as pure $z$-equation.
Alternative Method (for z ≠ 0)
$4z^2 + \bar{z} = 0$. For $z\neq 0$: multiply by $z$:
$4z^3 + z\bar{z} = 0 \implies 4z^3 + |z|^2 = 0$
Let $|z|^2 = r^2$: $z^3 = -r^2/4$
Also $|z^3| = r^3 \Rightarrow r^3 = r^2/4 \Rightarrow r = 1/4$ (for $r\neq 0$)
So $|z|=1/4$, $|z|^2=1/16$ for each non-zero solution ✓
3 non-zero solutions × $1/16$ = $3/16$ ✓
$4z^3 + z\bar{z} = 0 \implies 4z^3 + |z|^2 = 0$
Let $|z|^2 = r^2$: $z^3 = -r^2/4$
Also $|z^3| = r^3 \Rightarrow r^3 = r^2/4 \Rightarrow r = 1/4$ (for $r\neq 0$)
So $|z|=1/4$, $|z|^2=1/16$ for each non-zero solution ✓
3 non-zero solutions × $1/16$ = $3/16$ ✓
❓
Frequently Asked Questions
Q
How many solutions does 4z²+z̄=0 have?
▼
4 solutions: z=0 (trivial), z=−1/4 (real), z=1/8±i√3/8 (complex conjugate pair). Total |S|=4.
Q
Why does z=0 contribute 0 to the sum?
▼
|0|²=0 by definition. z=0 always satisfies 4(0)²+0̄=0, so it’s in S, but doesn’t affect the sum.
Q
Why do z₃ and z₄ have equal |z|²?
▼
z₃ and z₄ are complex conjugates (z₄=z̄₃). For conjugate pairs, |z|²=|z̄|² always. Both equal 1/64+3/64=1/16.
Q
What is the elegant approach using |z|²=zz̄?
▼
Multiply 4z²+z̄=0 by z: 4z³+|z|²=0. For z≠0: |z³|=|z|³=|z|²/4 → |z|=1/4 → |z|²=1/16. There are 3 non-zero solutions (cube roots), so Σ|z|²=3/16.
Q
How to verify z=1/8+i√3/8 satisfies 4z²+z̄=0?
▼
z²=(1/64−3/64)+2i(1/8)(√3/8)=(−1/32)+(i√3/32). So 4z²=−1/8+i√3/8. And z̄=1/8−i√3/8. Sum=−1/8+1/8+i(√3/8−√3/8)=0 ✓.
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