To solve $(-3 - 4i)^7$ and express it in $a + bi$ form, we can use De Moivre's theorem. Here are the steps:

1. Convert $(-3 - 4i)$ to polar form:

   • The magnitude $r$ is given by: $r = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5$
   • The angle $\theta$ is: $\theta = \tan^{-1}\left(\frac{-4}{-3}\right)$ Since the point $(-3, -4)$ is in the third quadrant, $\theta = \pi + \tan^{-1}\left(\frac{4}{3}\right)$.

2. Express in polar form: $-3 - 4i = 5 \text{cis}(\theta), \text{ where } \text{cis}(\theta) = \cos(\theta) + i\sin(\theta).$

3. Apply De Moivre's Theorem: For $(r \text{cis} \theta)^n$, it becomes: $r^n \text{cis}(n\theta).$ Here, $r = 5$, $n = 7$, and $\theta = \pi + \tan^{-1}\left(\frac{4}{3}\right)$: $\text{Magnitude: } 5^7 = 78125.$ $\text{Angle: } 7\theta = 7 \left(\pi + \tan^{-1}\left(\frac{4}{3}\right)\right).$

4. Simplify the trigonometric expressions: Use $\cos(7\theta)$ and $\sin(7\theta)$ to get the result in $a + bi$ form.

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