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.
I think I can read this as
z^7 = 1/√2 - 1/√2 i
so, that means
z^7 = cis 315°
z = cis(315/7)° = cis45° + k*360/7 for k=1..6
To find the roots of the equation $z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$, we can write the right-hand side in exponential form as follows:
\[-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \sqrt{2} \left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \sqrt{2} \left(\cos \left(-45^\circ\right) + i \sin \left(-45^\circ\right)\right)\]
By De Moivre's Theorem, the $n^\text{th}$ roots of a complex number in exponential form can be found by raising the magnitude to the power of $\frac{1}{n}$ and multiplying the argument by $\frac{1}{n}$. In this case, we seek the $7^\text{th}$ roots, so we have:
\begin{align*}
z &= \left(\sqrt{2}\right)^{\frac{1}{7}} \text{cis } \left(\frac{\text{Arg }\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)}{7}\right) \\
&= \sqrt[14]{2} \text{cis } \left(\frac{\text{Arg }\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)}{7}\right) \\
&= \sqrt[14]{2} \text{cis } \left(\frac{-45^\circ}{7}\right)
\end{align*}
Since we are looking for $7$ roots, we have $\theta_1, \theta_2, \dots, \theta_7$ such that $\theta_1 + \theta_2 + \dots + \theta_7 = 360^\circ$. Therefore, we need to find $\left(\frac{-45^\circ}{7}\right) + \left(\frac{-45^\circ}{7}\right) + \dots + \left(\frac{-45^\circ}{7}\right)$.
The sum of these angles can be found by multiplying the angle by $7$ since the desired sum is equal to the argument of $\left(z^7\right)^7$:
\begin{align*}
\left(\frac{-45^\circ}{7}\right) + \left(\frac{-45^\circ}{7}\right) + \dots + \left(\frac{-45^\circ}{7}\right) &= 7\left(\frac{-45^\circ}{7}\right) \\
&= -45^\circ
\end{align*}
Therefore, the value of $\theta_1 + \theta_2 + \dots + \theta_7$ is $\boxed{-45^\circ}$.
To find the values of $\theta_1, \theta_2, \dots, \theta_7$, we need to first rewrite the given complex number $-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ in polar form.
The polar form of a complex number $z = a + bi$ is given by $z = r\textrm{cis}(\theta)$, where $r$ is the magnitude (or modulus) of $z$ and $\theta$ is the argument of $z$.
In this case, we have $a = -\frac{1}{\sqrt{2}}$ and $b = -\frac{1}{\sqrt{2}}$. To find $r$ and $\theta$, we can use the following formulas:
\[r = \sqrt{a^2 + b^2}\]
\[\textrm{cis}(\theta) = \frac{a + bi}{r}\]
Plugging in the given values, we get:
\[r = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = 1\]
\[\textrm{cis}(\theta) = \frac{-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}}{1} = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]
So, the polar form of the complex number $-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ is $1\textrm{cis}(\theta)$, where $\theta$ is the argument of $-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$.
To find the argument (or angle) $\theta$, we can use the inverse tangent function:
\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]
Plugging in the given values, we get:
\[\theta = \tan^{-1}\left(\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}\right) = \tan^{-1}(1) = 45^\circ\]
Therefore, the complex number $-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ can be written as $1\textrm{cis}(45^\circ)$.
Now, since we are given that $z^7 = 1\textrm{cis}(45^\circ)$, we can find the 7th roots by using De Moivre's Theorem, which states that if $z = r\textrm{cis}(\theta)$, then the $n$th roots of $z$ are given by:
\[z^{\frac{1}{n}} = r^\frac{1}{n}\textrm{cis}\left(\frac{\theta + 2k\pi}{n}\right)\]
where $k$ is an integer.
In this case, we have $z = 1\textrm{cis}(45^\circ)$, so the $7$th roots of $z$ are given by:
\[z^{\frac{1}{7}} = 1^\frac{1}{7}\textrm{cis}\left(\frac{45^\circ + 2k\pi}{7}\right)\]
for $k = 0,1,2,\dots,6$.
We can now find the values of $\theta_1, \theta_2, \dots, \theta_7$ by plugging in the different values of $k$:
For $k = 0$: $\theta_1 = \frac{45^\circ + 2(0)\pi}{7} = \frac{45^\circ}{7}$
For $k = 1$: $\theta_2 = \frac{45^\circ + 2(1)\pi}{7} = \frac{45^\circ + 2\pi}{7}$
For $k = 2$: $\theta_3 = \frac{45^\circ + 2(2)\pi}{7} = \frac{45^\circ + 4\pi}{7}$
For $k = 3$: $\theta_4 = \frac{45^\circ + 2(3)\pi}{7} = \frac{45^\circ + 6\pi}{7}$
For $k = 4$: $\theta_5 = \frac{45^\circ + 2(4)\pi}{7} = \frac{45^\circ + 8\pi}{7}$
For $k = 5$: $\theta_6 = \frac{45^\circ + 2(5)\pi}{7} = \frac{45^\circ + 10\pi}{7}$
For $k = 6$: $\theta_7 = \frac{45^\circ + 2(6)\pi}{7} = \frac{45^\circ + 12\pi}{7}$
Now, we can find $\theta_1 + \theta_2 + \dots + \theta_7$:
\[\theta_1 + \theta_2 + \dots + \theta_7 = \frac{45^\circ}{7} + \frac{45^\circ + 2\pi}{7} + \frac{45^\circ + 4\pi}{7} + \frac{45^\circ + 6\pi}{7} + \frac{45^\circ + 8\pi}{7} + \frac{45^\circ + 10\pi}{7} + \frac{45^\circ + 12\pi}{7}\]
Simplifying this expression, we can combine the constants ($45^\circ$) and the terms with $\pi$:
\[\theta_1 + \theta_2 + \dots + \theta_7 = \frac{7 \cdot 45^\circ + 2\pi + 4\pi + 6\pi + 8\pi + 10\pi + 12\pi}{7}\]
\[\theta_1 + \theta_2 + \dots + \theta_7 = \frac{7 \cdot 45^\circ + (2+4+6+8+10+12)\pi}{7}\]
\[\theta_1 + \theta_2 + \dots + \theta_7 = \frac{7 \cdot 45^\circ + 42\pi}{7}\]
Simplifying further,
\[\theta_1 + \theta_2 + \dots + \theta_7 = 45^\circ + 6\pi\]
Therefore, the sum of $\theta_1, \theta_2, \dots, \theta_7$ is $45^\circ + 6\pi$ degrees.