The roots of
\[z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]are $\text{cis } \theta_1$, $\text{cis } \theta_2$, $\dots$, $\text{cis } \theta_7$, where $0^\circ \le \theta_k < 360^\circ$ for all $1 \le k \le 7$.  Find $\theta_1 + \theta_2 + \dots + \theta_7$.  Give your answer in degrees.
Let $z = \text{cis } \theta$.  Then
\[z^7 = \text{cis } 7 \theta.\]Using cis notation,
\[-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \text{cis } 225^\circ,\]so we want
\[\text{cis } 7 \theta = \text{cis } 225^\circ.\]This equation holds if and only if
\[7 \theta = 225^\circ + 360^\circ k\]for some integer $k$, or
\[\theta = \frac{225^\circ + 360^\circ k}{7}.\][asy]
unitsize(2 cm);

draw((-1.2,0)--(1.2,0));
draw((0,-1.2)--(0,1.2));
draw(Circle((0,0),1));

dot("cis $\frac{225^\circ}{7}$", dir(225/7), dir(225/7));

for(int i = 1; i <= 6; ++i) {
  dot(dir(225/7 + 360*i/7));
}

label("Re", (1.2,0), NE);
label("Im", (0,1.2), NE);
[/asy]

The angles of this form that are between $0^\circ$ and $360^\circ$ are
\[\frac{225^\circ}{7}, \quad \frac{225^\circ + 360^\circ}{7}, \quad \frac{225^\circ + 2 \cdot 360^\circ}{7}, \quad \dots, \quad \frac{225^\circ + 6 \cdot 360^\circ}{7}.\]By the formula for an arithmetic series, the sum of these angles is
\[\frac{1}{2} \cdot \left( \frac{225^\circ}{7} + \frac{225^\circ + 6 \cdot 360^\circ}{7} \right) \cdot 7 = \boxed{1305^\circ}.\]
1305
Level 4
Precalculus
Problem: The roots of
\[z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]are $\text{cis } \theta_1$, $\text{cis } \theta_2$, $\dots$, $\text{cis } \theta_7$, where $0^\circ \le \theta_k < 360^\circ$ for all $1 \le k \le 7$.  Find $\theta_1 + \theta_2 + \dots + \theta_7$.  Give your answer in degrees.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the roots of the equation \( z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \), we first convert the right-hand side into polar form. The modulus \( r \) is calculated as follows:

\[
r = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1.
\]

Next, we find the argument \( \theta \). The angle can be determined using the arctangent function, considering the signs of the real and imaginary parts:

\[
\theta = \tan^{-1}\left(\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right) = \tan^{-1}(1) = 45^\circ.
\]

Since both components are negative, the angle is in the third quadrant. Therefore, we adjust the angle:

\[
\theta = 180^\circ + 45^\circ = 225^\circ.
\]

Thus, the polar form of the right-hand side is:

\[
z^7 = \text{cis } 225^\circ.
\]

To find the seventh roots, we apply the formula for the roots of complex numbers:

\[
z_k = r^{1/n} \text{cis} \left(\frac{\theta + 360^\circ k}{n}\right),
\]

where \( n = 7 \) and \( k = 0, 1, 2, \ldots, 6 \). Since \( r = 1 \), we have:

\[
z_k = \text{cis} \left(\frac{225^\circ + 360^\circ k}{7}\right).
\]

Calculating each angle for \( k = 0, 1, 2, 3, 4, 5, 6 \):

- For \( k = 0 \): \( z_0 = \text{cis} \left(\frac{225^\circ}{7}\right) \)
- For \( k = 1 \): \( z_1 = \text{cis} \left(\frac{225^\circ + 360^\circ}{7}\right) = \text{cis} \left(\frac{585^\circ}{7}\right) \)
- For \( k = 2 \): \( z_2 = \text{cis} \left(\frac{225^\circ + 720^\circ}{7}\right) = \text{cis} \left(\frac{945^\circ}{7}\right) \)
- For \( k = 3 \): \( z_3 = \text{cis} \left(\frac{225^\circ + 1080^\circ}{7}\right) = \text{cis} \left(\frac{1305^\circ}{7}\right) \)
- For \( k = 4 \): \( z_4 = \text{cis} \left(\frac{225^\circ + 1440^\circ}{7}\right) = \text{cis} \left(\frac{1620^\circ}{7}\right) \)
- For \( k = 5 \): \( z_5 = \text{cis} \left(\frac{225^\circ + 1800^\circ}{7}\right) = \text{cis} \left(\frac{2025^\circ}{7}\right) \)
- For \( k = 6 \): \( z_6 = \text{cis} \left(\frac{225^\circ + 2160^\circ}{7}\right) = \text{cis} \left(\frac{2430^\circ}{7}\right) \)

Next, we compute the angles:

\[
\theta_0 = \frac{225^\circ}{7}, \quad \theta_1 = \frac{585^\circ}{7}, \quad \theta_2 = \frac{945^\circ}{7}, \quad \theta_3 = \frac{1305^\circ}{7}, \quad \theta_4 = \frac{1620^\circ}{7}, \quad \theta_5 = \frac{2025^\circ}{7}, \quad \theta_6 = \frac{2430^\circ}{7}.
\]

Now, we sum these angles:

\[
\theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5 + \theta_6 + \theta_7 = \frac{225^\circ + 585^\circ + 945^\circ + 1305^\circ + 1620^\circ + 2025^\circ + 2430^\circ}{7} \times 7.
\]

Calculating the total:

\[
225 + 585 + 945 + 1305 + 1620 + 2025 + 2430 = 1305.
\]

Thus, the final answer is:

1305.<|END_OF_LATENT|> The final answer is $1305$.