find all the complex roots. write the answer in exponential form.\nthe complex cube roots of $12 +…

Explanation:

Step1: Convert to polar form

First, find the modulus $r$ of $12+12i$: $r = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2}$ Find the argument $\theta$: $\theta = \arctan\left(\frac{12}{12}\right) = \frac{\pi}{4}$ Polar form: $12\sqrt{2}e^{i\left(\frac{\pi}{4} + 2\pi k\right)}$, $k=0,1,2$

Step2: Apply root formula

For $n$-th roots, use $r^{\frac{1}{n}}e^{i\left(\frac{\theta + 2\pi k}{n}\right)}$, $n=3$ Modulus of roots: $(12\sqrt{2})^{\frac{1}{3}} = (12)^{\frac{1}{3}}(2^{\frac{1}{2}})^{\frac{1}{3}} = 12^{\frac{1}{3}}2^{\frac{1}{6}} = (2^2 \cdot 3)^{\frac{1}{3}}2^{\frac{1}{6}} = 2^{\frac{5}{6}}3^{\frac{1}{3}}$ Simplify $2^{\frac{5}{6}}3^{\frac{1}{3}} = (2^5 \cdot 3^2)^{\frac{1}{6}} = (32 \cdot 9)^{\frac{1}{6}} = 288^{\frac{1}{6}}$, or keep as $12^{\frac{1}{3}}e^{i\frac{\pi + 8\pi k}{12}}$

Step3: Calculate each root

For $k=0$: $288^{\frac{1}{6}}e^{i\frac{\pi}{12}}$ For $k=1$: $288^{\frac{1}{6}}e^{i\left(\frac{\pi + 8\pi}{12}\right)} = 288^{\frac{1}{6}}e^{i\frac{3\pi}{4}}$ For $k=2$: $288^{\frac{1}{6}}e^{i\left(\frac{\pi + 16\pi}{12}\right)} = 288^{\frac{1}{6}}e^{i\frac{17\pi}{12}}$

Answer:

$288^{\frac{1}{6}}e^{i\frac{\pi}{12}}, 288^{\frac{1}{6}}e^{i\frac{3\pi}{4}}, 288^{\frac{1}{6}}e^{i\frac{17\pi}{12}}$