The number of roots of the equation `z^(6)=-64` whose real parts are non-negative,

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the equation \( z^6 = -64 \) and find the number of roots whose real parts are non-negative, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ z^6 = -64 \]
This can be rewritten as:
\[ z^6 = 64 \cdot e^{i\pi} \]
Here, \( -64 \) is represented in polar form, where \( 64 \) is the modulus and \( \pi \) is the argument (angle).

### Step 2: Find the modulus and argument
The modulus of \( -64 \) is \( 64 \), and the argument is \( \pi \). Therefore, we can express \( z \) in polar form:
\[ z = 64^{1/6} \cdot e^{i(\pi + 2k\pi)/6} \]
where \( k \) is any integer.

### Step 3: Calculate the sixth root of the modulus
Calculating \( 64^{1/6} \):
\[ 64^{1/6} = (2^6)^{1/6} = 2 \]
Thus, we have:
\[ z = 2 \cdot e^{i(\pi/6 + k\pi/3)} \]

### Step 4: Determine the angles for the roots
Now we can find the angles for \( k = 0, 1, 2, 3, 4, 5 \):
- For \( k = 0 \): \( z = 2 \cdot e^{i\pi/6} \)
- For \( k = 1 \): \( z = 2 \cdot e^{i(\pi/6 + \pi/3)} = 2 \cdot e^{i\pi/2} \)
- For \( k = 2 \): \( z = 2 \cdot e^{i(\pi/6 + 2\pi/3)} = 2 \cdot e^{i(5\pi/6)} \)
- For \( k = 3 \): \( z = 2 \cdot e^{i(\pi/6 + \pi)} = 2 \cdot e^{i(7\pi/6)} \)
- For \( k = 4 \): \( z = 2 \cdot e^{i(\pi/6 + 4\pi/3)} = 2 \cdot e^{i(3\pi/2)} \)
- For \( k = 5 \): \( z = 2 \cdot e^{i(\pi/6 + 5\pi/3)} = 2 \cdot e^{i(11\pi/6)} \)

### Step 5: Identify the roots with non-negative real parts
Next, we convert these angles to Cartesian coordinates and check the real parts:
1. \( z_0 = 2 \cdot \left( \cos(\pi/6) + i \sin(\pi/6) \right) = 2 \cdot \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \sqrt{3} + i \)
2. \( z_1 = 2 \cdot \left( \cos(\pi/2) + i \sin(\pi/2) \right) = 2 \cdot (0 + i) = 2i \)
3. \( z_2 = 2 \cdot \left( \cos(5\pi/6) + i \sin(5\pi/6) \right) = 2 \cdot \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i \)
4. \( z_3 = 2 \cdot \left( \cos(7\pi/6) + i \sin(7\pi/6) \right) = 2 \cdot \left( -\frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = -\sqrt{3} - i \)
5. \( z_4 = 2 \cdot \left( \cos(3\pi/2) + i \sin(3\pi/2) \right) = 2 \cdot (0 - i) = -2i \)
6. \( z_5 = 2 \cdot \left( \cos(11\pi/6) + i \sin(11\pi/6) \right) = 2 \cdot \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = \sqrt{3} - i \)

From these, the roots with non-negative real parts are:
- \( z_0 = \sqrt{3} + i \) (real part positive)
- \( z_1 = 2i \) (real part zero)
- \( z_2 = -\sqrt{3} + i \) (real part negative)
- \( z_3 = -\sqrt{3} - i \) (real part negative)
- \( z_4 = -2i \) (real part zero)
- \( z_5 = \sqrt{3} - i \) (real part positive)

### Step 6: Count the roots with non-negative real parts
The roots with non-negative real parts are \( z_0, z_1, z_5 \). Thus, there are a total of **4 roots** with non-negative real parts.

### Final Answer
The number of roots of the equation \( z^6 = -64 \) whose real parts are non-negative is **4**.

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