Let 'z' be a complex number and 'a' be a real parameter such that `z^2+az+a^2=0`, then

Options

A. locus of z is a pair of straight lines
B. locus of z is a circle
C. `"arg" (z) = pm (2pi)/(3)`
D. `|z|= |a|`

Answer

To solve the equation \( z^2 + az + a^2 = 0 \) where \( z \) is a complex number and \( a \) is a real parameter, we can follow these steps:

### Step 1: Identify the coefficients
In the quadratic equation \( z^2 + az + a^2 = 0 \), we identify the coefficients:
- \( A = 1 \)
- \( B = a \)
- \( C = a^2 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = a^2 - 4(1)(a^2) = a^2 - 4a^2 = -3a^2 \]

### Step 3: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ z = \frac{-B \pm \sqrt{D}}{2A} \]
Substituting \( B = a \), \( D = -3a^2 \), and \( A = 1 \):
\[ z = \frac{-a \pm \sqrt{-3a^2}}{2} \]
Since \( \sqrt{-3a^2} = i\sqrt{3}a \), we can rewrite it as:
\[ z = \frac{-a \pm i\sqrt{3}a}{2} \]

### Step 4: Simplify the expression
Now, we can separate the real and imaginary parts:
\[ z = \frac{-a}{2} \pm \frac{i\sqrt{3}a}{2} \]
This gives us two values for \( z \):
\[ z_1 = \frac{-a + i\sqrt{3}a}{2}, \quad z_2 = \frac{-a - i\sqrt{3}a}{2} \]

### Step 5: Find the modulus of \( z \)
The modulus of \( z \) is given by:
\[ |z| = \sqrt{\left(\frac{-a}{2}\right)^2 + \left(\frac{\sqrt{3}a}{2}\right)^2} \]
Calculating this:
\[ |z| = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{\frac{4a^2}{4}} = \sqrt{a^2} = |a| \]

### Step 6: Find the argument of \( z \)
The argument of \( z \) can be found using:
\[ \text{arg}(z) = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{3}a}{2}}{\frac{-a}{2}}\right) = \tan^{-1}(-\sqrt{3}) \]
This corresponds to angles in the second and fourth quadrants. The principal value of \( \tan^{-1}(-\sqrt{3}) \) is \( -\frac{\pi}{3} \), which can be adjusted to the second quadrant:
\[ \text{arg}(z) = \frac{2\pi}{3} \text{ or } -\frac{\pi}{3} \]

### Final Result
Thus, we conclude that:
- The modulus of \( z \) is \( |a| \).
- The argument of \( z \) is \( \frac{2\pi}{3} \) or \( -\frac{\pi}{3} \).