Given the complex number $a=\cos\theta+i\sin\theta$. We need to find the value of $1+a^2$.

Finding $a^2$

Using the property of complex numbers in polar form (De Moivre's theorem), if $a = \cos\theta + i\sin\theta$, then $a^n = \cos(n\theta) + i\sin(n\theta)$.

Therefore, for $n=2$:

$a^2 = \cos(2\theta) + i\sin(2\theta)$

Calculating $1+a^2$

Now, substitute the expression for $a^2$ into $1+a^2$:

$1+a^2 = 1 + (\cos(2\theta) + i\sin(2\theta))$

$1+a^2 = (1 + \cos(2\theta)) + i\sin(2\theta)$

Applying Trigonometric Identities

Use the double angle identities:

• $1 + \cos(2\theta) = 2\cos^2\theta$
• $\sin(2\theta) = 2\sin\theta\cos\theta$

Substitute these identities back into the expression for $1+a^2$:

$1+a^2 = (2\cos^2\theta) + i(2\sin\theta\cos\theta)$

Factoring the Expression

Factor out the common term $2\cos\theta$ from both the real and imaginary parts:

$1+a^2 = 2\cos\theta(\cos\theta + i\sin\theta)$

This matches the expression given in Option A.