To find the modulus of the complex number $1 + \cos \alpha + i \sin \alpha$, we can use trigonometric identities.

Finding the Modulus of 1 + cos α + i sin α

Let the complex number be $z = 1 + \cos \alpha + i \sin \alpha$. We can rewrite this using the following trigonometric identities:

• $1 + \cos \alpha = 2\cos^2 \frac{\alpha}{2}$
• $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$

Substitute these identities into the expression for $z$:

$$ z = \left( 2\cos^2 \frac{\alpha}{2} \right) + i \left( 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \right) $$

Now, we can factor out the common term $2\cos \frac{\alpha}{2}$ from both the real and imaginary parts:

$$ z = 2\cos \frac{\alpha}{2} \left( \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right) $$

Calculating the Modulus

The modulus of a complex number $a+bi$ is given by $|a+bi| = \sqrt{a^2 + b^2}$. Alternatively, we can use the property that for a product of complex numbers (or a complex number and a real number), the modulus of the product is the product of the moduli: $|zw| = |z||w|$.

In our case, $z = 2\cos \frac{\alpha}{2} \times \left( \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right)$.

Let's find the modulus of each part:

• The modulus of the first part is $|2\cos \frac{\alpha}{2}|$.
• The modulus of the second part, $\left( \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right)$, can be calculated as: $$ \left| \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right| = \sqrt{\cos^2 \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2}} $$ Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we get: $$ \sqrt{\cos^2 \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2}} = \sqrt{1} = 1 $$

Therefore, the modulus of $z$ is:

$$ |z| = \left| 2\cos \frac{\alpha}{2} \right| \times 1 = \left| 2\cos \frac{\alpha}{2} \right| $$

Typically, when trigonometric functions are involved in modulus calculations like this, and the options don't include absolute value signs, the expression is often simplified assuming $\cos \frac{\alpha}{2}$ is non-negative (e.g., for $\alpha$ in the range $[-\pi, \pi]$). Under this assumption, the modulus is $2\cos \frac{\alpha}{2}$.

Comparing this with the given options, the modulus is $2\cos \frac{\alpha}{2}$.

Final Answer Check

The calculated modulus is $2\cos \frac{\alpha}{2}$, which corresponds to option 2.