Express the complex numbers modulus -amplitude form
`-1-isqrt(3)`

Answer

To express the complex number \(-1 - i\sqrt{3}\) in modulus-amplitude form, we will follow these steps:

### Step 1: Identify the complex number
The given complex number is:
\[ z = -1 - i\sqrt{3} \]

### Step 2: Calculate the modulus
The modulus \(r\) of a complex number \(z = a + bi\) is given by:
\[ r = \sqrt{a^2 + b^2} \]
Here, \(a = -1\) and \(b = -\sqrt{3}\).

Calculating the modulus:
\[ r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 3: Determine the amplitude (angle)
To find the amplitude \(\theta\), we use the relationships:
\[ \cos \theta = \frac{a}{r} \quad \text{and} \quad \sin \theta = \frac{b}{r} \]
Substituting the values we have:
\[ \cos \theta = \frac{-1}{2} \quad \text{and} \quad \sin \theta = \frac{-\sqrt{3}}{2} \]

### Step 4: Identify the quadrant
Since both cosine and sine are negative, the angle \(\theta\) must be in the third quadrant.

### Step 5: Find the reference angle
The reference angle where \(\cos \theta = \frac{1}{2}\) and \(\sin \theta = \frac{\sqrt{3}}{2}\) is \(60^\circ\). Therefore, in the third quadrant, the angle is:
\[ \theta = 180^\circ + 60^\circ = 240^\circ \]

### Step 6: Write the modulus-amplitude form
Now we can express the complex number in modulus-amplitude form:
\[ z = r(\cos \theta + i\sin \theta) = 2\left(\cos 240^\circ + i\sin 240^\circ\right) \]

### Final Answer
Thus, the modulus-amplitude form of the complex number \(-1 - i\sqrt{3}\) is:
\[ z = 2\left(\cos 240^\circ + i\sin 240^\circ\right) \]

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