I assume the OP means a Complex normal distribution whose covariance and relation matrices are identity matrices. To get at the result we use a few properties; first $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, second that diagonal covariance and relation matrices imply independence of random variables, third is the geometric series.

Start with:

$$\mathbb{E}(\phi(k)) = \sqrt{1-\rho^2}\sum_{j=1}^{k-1}\mathbb{E}(\rho^{k-j-1}e(j)) $$

and notice that the $\rho$ is constant with respect to the expectation so that each term in the summand is multiplied by $0$. Therefore, $\mathbb{E}(\phi(k)) = 0$.

Next look at the second moment:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\left(\sum_{j=1}^{k-1}\rho^{k-j-1}e(j)\right)^2\right) $$ which, expanding the squares, simplifies to:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2 + \sum_{j=1, j\neq i}^{k-1}\sum_{i=1}^{k-1} \rho^{2k-i-j-2} e(i)e(j)\right) $$

Now putting the expectation operator through to the two terms of the square shows that the second term in the sum is $0$, because all terms in the double sum are multiplied by a $0$ term. The first term in the sum has:

$$\mathbb{E}(e(i)^2) = \sigma^2 $$

So that the formula simplifies:

$$ (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2\right) = (1-\rho^2)\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}\sigma^2 \right)$$

Next factor out the common $\sigma^2$ term and identify the sum as (part of) a geometric series (missing the $j=0$ term. Using the formula

$$\sum_{j=0}^{k-1} (\rho^2)^j = \frac{1-\rho^{2k}}{1-\rho^2}$$

we simplify to

$$ (1 - \rho^2) \sigma^2 \left( \sum_{j=1}^{k-1}\rho^{2(k-j-1)} \right) = (1-\rho^2)\sigma^2 \rho^{2(k-1)} \left( \frac{1-\rho^{2k}}{1-\rho^2} - 1 \right). $$

This last term on the right hand side is what you'd likely use (or some variation). For example, you could turn the $-1$ term into a common denominator and factor out a $\rho^2$ if you like too but this general approach.

I assume the OP means a Complex normal distribution whose covariance and relation matrices are identity matrices. To get at the result we use a few properties; first $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, second that diagonal covariance and relation matrices imply independence of random variables, third is the geometric series.

Start with:

$$\mathbb{E}(\phi(k)) = \sqrt{1-\rho^2}\sum_{j=1}^{k-1}\mathbb{E}(\rho^{k-j-1}e(j)) $$

and notice that the $\rho$ is constant with respect to the expectation so that each term in the summand is multiplied by $0$. Therefore, $\mathbb{E}(\phi(k)) = 0$.

Next look at the second moment:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\left(\sum_{j=1}^{k-1}\rho^{k-j-1}e(j)\right)^2\right) $$ which, expanding the squares, simplifies to:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2 + \sum_{j=1, j\neq i}^{k-1}\sum_{i=1}^{k-1} \rho^{2k-i-j-2} e(i)e(j)\right) $$

Now putting the expectation operator through to the two terms of the square shows that the second term in the sum is $0$, because all terms in the double sum are multiplied by a $0$ term. The first term in the sum has:

$$\mathbb{E}(e(i)^2) = \sigma^2 $$

So that the formula simplifies:

$$ (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2\right) = (1-\rho^2)\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}\sigma^2 \right)$$

Next factor out the common $\sigma^2$ term and identify the sum as (part of) a geometric series (missing the $j=0$ term. Using the formula

$$\sum_{j=0}^{k-1} (\rho^2)^j = \frac{1-\rho^{2k}}{1-\rho^2}$$

we simplify to

$$ (1 - \rho^2) \sigma^2 \left( \sum_{j=1}^{k-1}\rho^{2(k-j-1)} \right) = (1-\rho^2)\sigma^2 \rho^{2(k-1)} \left( \frac{1-\rho^{2k}}{1-\rho^2} - 1 \right). $$

This last term on the right hand side is what you'd likely use (or some variation). For example, you could turn the $-1$ term into a common denominator and factor out a $\rho^2$ if you like too but this general approach.

Also, here we are assuming $k \geq 1$.

I assume the OP means a Complex normal distribution whose covariance and relation matrices are identity matrices. To get at the result we use a few properties; first $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, second that diagonal covariance and relation matrices imply independence of random variables, third is the geometric series.

Start with:

$$\mathbb{E}(\phi(k)) = \sqrt{1-\rho^2}\sum_{j=1}^{k-1}\mathbb{E}(\rho^{k-j-1}e(j)) $$

and notice that the $\rho$ is constant with respect to the expectation so that each term in the summand is multiplied by $0$. Therefore, $\mathbb{E}(\phi(k)) = 0$.

Next look at the second moment:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\left(\sum_{j=1}^{k-1}\rho^{k-j-1}e(j)\right)^2\right) $$ which, expanding the squares, simplifies to:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2 + \sum_{j=1, j\neq i}^{k-1}\sum_{i=1}^{k-1} \rho^{2k-i-j-2} e(i)e(j)\right) $$

Now putting the expectation operator through to the two terms of the square shows that the second term in the sum is $0$, because all terms in the double sum are multiplied by a $0$ term. The first term in the sum has:

$$\mathbb{E}(e(i)^2) = \sigma^2 $$

So that the formula simplifies:

$$ (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2\right) = (1-\rho^2)\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}\sigma^2 \right)$$

Next factor out the common $\sigma^2$ term and identify the sum as (part of) a geometric series (missing the $j=0$ term. Using the formula

$$\sum_{j=0}^{k-1} (\rho^2)^j = \frac{1-\rho^{2k}}{1-\rho^2}$$

we simplify to

$$ (1 - \rho^2) \sigma^2 \left( \sum_{j=1}^{k-1}\rho^{2(k-j-1)} \right) = (1-\rho^2)\sigma^2 \left( \frac{1-\rho^{2k}}{1-\rho^2} - 1 \right). $$

This last term on the right hand side is what you'd likely use (or some variation). For example, you could turn the $-1$ term into a common denominator and factor out a $\rho^2$ if you like too but this general approach.

I assume the OP means a Complex normal distribution whose covariance and relation matrices are identity matrices. To get at the result we use a few properties; first $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, second that diagonal covariance and relation matrices imply independence of random variables, third is the geometric series.

Start with:

$$\mathbb{E}(\phi(k)) = \sqrt{1-\rho^2}\sum_{j=1}^{k-1}\mathbb{E}(\rho^{k-j-1}e(j)) $$

and notice that the $\rho$ is constant with respect to the expectation so that each term in the summand is multiplied by $0$. Therefore, $\mathbb{E}(\phi(k)) = 0$.

Next look at the second moment:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\left(\sum_{j=1}^{k-1}\rho^{k-j-1}e(j)\right)^2\right) $$ which, expanding the squares, simplifies to:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2 + \sum_{j=1, j\neq i}^{k-1}\sum_{i=1}^{k-1} \rho^{2k-i-j-2} e(i)e(j)\right) $$

Now putting the expectation operator through to the two terms of the square shows that the second term in the sum is $0$, because all terms in the double sum are multiplied by a $0$ term. The first term in the sum has:

$$\mathbb{E}(e(i)^2) = \sigma^2 $$

So that the formula simplifies:

$$ (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2\right) = (1-\rho^2)\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}\sigma^2 \right)$$

Next factor out the common $\sigma^2$ term and identify the sum as (part of) a geometric series (missing the $j=0$ term. Using the formula

$$\sum_{j=0}^{k-1} (\rho^2)^j = \frac{1-\rho^{2k}}{1-\rho^2}$$

we simplify to

$$ (1 - \rho^2) \sigma^2 \left( \sum_{j=1}^{k-1}\rho^{2(k-j-1)} \right) = (1-\rho^2)\sigma^2 \rho^{2(k-1)} \left( \frac{1-\rho^{2k}}{1-\rho^2} - 1 \right). $$

This last term on the right hand side is what you'd likely use (or some variation). For example, you could turn the $-1$ term into a common denominator and factor out a $\rho^2$ if you like too but this general approach.