$$ 6 \sqrt{2}-6 \sqrt{2} i $$ The square roots, in trigonometric notation, are $$ 2 \sqrt{3}(\cos 157.5+i \sin 157.5), 2 \sqrt{3}(\cos 337.5+i \sin 337.5) $$. (Type an exact answer, using radicals and $$ i $$ as needed. Type any angle measures in degrees. Use angle measures greater than or equal to 0 and less than 360 . Do not include the degree symbol in your answer. Use integers or decimals for any numbers in the expression. Use a comma to separate answers as needed.) Try again. First write the complex number in trigonometric notation. Then, solve for the square roots of the number. The nth roots of a complex number $$ r(\cos heta+i \sin heta) $$, $$ r
eq 0 $$, are given by the following formula, where $$ k=0,1,2, \ldots, n-1 $$. $$ r^{1 / n}\left[\cos \left(\frac{ heta}{n}+k \cdot \frac{360^{\circ}}{n} ight)+i \sin \left(\frac{ heta}{n}+k \cdot \frac{360^{\circ}}{n} ight) ight] $$

Question

$$ 6 \sqrt{2}-6 \sqrt{2} i $$ The square roots, in trigonometric notation, are $$ 2 \sqrt{3}(\cos 157.5+i \sin 157.5), 2 \sqrt{3}(\cos 337.5+i \sin 337.5) $$. (Type an exact answer, using radicals and $$ i $$ as needed. Type any angle measures in degrees. Use angle measures greater than or equal to 0 and less than 360 . Do not include the degree symbol in your answer. Use integers or decimals for any numbers in the expression. Use a comma to separate answers as needed.) Try again. First write the complex number in trigonometric notation. Then, solve for the square roots of the number. The nth roots of a complex number $$ r(\cos 	heta+i \sin 	heta) $$, $$ r 
eq 0 $$, are given by the following formula, where $$ k=0,1,2, \ldots, n-1 $$. $$ r^{1 / n}\left[\cos \left(\frac{	heta}{n}+k \cdot \frac{360^{\circ}}{n}ight)+i \sin \left(\frac{	heta}{n}+k \cdot \frac{360^{\circ}}{n}ight)ight] $$

UpStudy AI · Accepted Answer

We start with the complex number
$$
6\sqrt{2}-6\sqrt{2} i.
$$
We first write it in trigonometric form.

1. Compute the modulus $$ r $$:
   $$
   r=\sqrt{(6\sqrt{2})^2+(-6\sqrt{2})^2}=\sqrt{72+72}=\sqrt{144}=12.
   $$

2. Find the argument $$\theta$$. Since the real part is positive and the imaginary part is negative,
   $$
   \theta=\arctan\frac{-6\sqrt{2}}{6\sqrt{2}}=\arctan(-1)=-45.
   $$
   To express the angle in the interval $$[0,360)$$ we add $$360$$:
   $$
   \theta=360-45=315.
   $$

Thus, the complex number in trigonometric form is
$$
12\left[\cos315+i\sin315\right].
$$

Now we find its square roots. In general, the $$n$$th roots of a complex number 
$$
r(\cos\theta+i\sin\theta)
$$
are given by
$$
r^{1/n}\left[\cos\left(\frac{\theta}{n}+k\cdot\frac{360}{n}\right)+i\sin\left(\frac{\theta}{n}+k\cdot\frac{360}{n}\right)\right],\quad k=0,1,\ldots, n-1.
$$

For the square roots $$(n=2)$$:

3. Compute the square root of the modulus:
   $$
   \sqrt{12}=2\sqrt{3}.
   $$

4. The two roots are given by:
   - For $$k=0$$:
     $$
     \alpha_0=2\sqrt{3}\left[\cos\left(\frac{315}{2}\right)+i\sin\left(\frac{315}{2}\right)\right]
     =2\sqrt{3}\left(\cos157.5+i\sin157.5\right).
     $$
   - For $$k=1$$:
     $$
     \alpha_1=2\sqrt{3}\left[\cos\left(\frac{315}{2}+180\right)+i\sin\left(\frac{315}{2}+180\right)\right]
     =2\sqrt{3}\left(\cos337.5+i\sin337.5\right).
     $$

Thus, the square roots of $$6\sqrt{2}-6\sqrt{2} i$$ are:
$$
2\sqrt{3}\left(\cos157.5+i\sin157.5\right),\quad 2\sqrt{3}\left(\cos337.5+i\sin337.5\right).
$$
The square roots of $$6\sqrt{2}-6\sqrt{2} i$$ are: $$ 2\sqrt{3}(\cos157.5 + i\sin157.5),\quad 2\sqrt{3}(\cos337.5 + i\sin337.5) $$

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