Determine the fourth roots of -16 in the form $x +iy$ where $x$ and $y$ are not trigonometric functions.
I do not even know what they really want from me in this question.
My initial thought wass: $\sqrt[4]{-16} = 0 + 2i = 2i$
but that seems overly simplistic for a problem that counts as 8 points in an exam.
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$\begingroup$Your equation is $z^4=-16$. Notice that $-16=(4i)^2$, so $$ z^4+16=(z^2-4i)(z^2+4i) $$ and the equation reduces to solve $z^2-4i=0$ and $z^2+4i=0$, that is, finding the square roots of $4i$ and $-4i$, which should be easy: write $$ (x+iy)^2=4i $$ and compare the real and imaginary parts. Do the same for $-4i$.
$\endgroup$ $\begingroup$In the complex numbers, you can't just take the $n$th root of a number and get one answer. There are always $2$ square roots, $3$ cube roots, $4$ fourth roots, etc. It's a different world than the real numbers.
Anyway, the way they likely want you to solve this is to use the polar form $-16 = 16e^{\pi i}$. $16$ is the radius and $\pi$ is the angle. To find the $n$th roots of a number in this form, you have to do two things
- Take the $n$th root of the radius.
- Divide the angle by $n$ and then add all possible multiples of $\frac{2\pi}{n}$. (You should get $n$ different angles.)
Here, you take the fourth root of $16$, which is $2$. Then dividing the angle by $4$ gives you $\frac{\pi}{4}$, and adding all possible multiples of $\frac{2\pi}{4} = \frac{\pi}{2}$ gives you $\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, $\frac{7\pi}{4}$. This ives you four answers: $$ 2e^{\frac{\pi}{4} i}, 2e^{\frac{3\pi}{4} i}, 2e^{\frac{5\pi}{4} i}, 2e^{\frac{7\pi}{4} i} $$ You can leave these in polar form $re^{\theta i}$ or you can convert them to the form $a + bi$ using the formula $e^{ix} = \cos x + i\sin x$.
$\endgroup$ $\begingroup$First make a substitution: $i^2 = -1$
$$\sqrt[4]{-16} = \sqrt[4]{16i^2} = \pm2\sqrt{i}$$
Now we have to find the square root of of i. We have:
$$(a+bi)^2 = i$$ $$(a^2-b^2) + (2ab)i = 0 + 1i$$
This implies that $a^2-b^2 = 0$ We have 2 options: $a=-b$ and $a=b$
For first option we have $-2b^2 = 1$, which doesnt have a solution in $\mathbb{R}$ It means that $a=b$. This implies:
$$2b^2 = 1$$ $$b = a = \pm\frac{1}{\sqrt{2}}$$
So substituting back we have:
$$\sqrt{i} = \pm\left(\frac{1}{\sqrt{2}}\right)(1+i)$$
So this leads to 2 solutions:
$$\sqrt[4]{-16} = \pm\left(\frac{2}{\sqrt{2}}\right)(1+i)$$
And your thought is wrong because:
$$(2i)^4 = 16i^4 = 16\text{, not $-16$}$$
$\endgroup$ $\begingroup$All right, what you first have to do in a problem like this is convert the statement -16 into polar coordinates, so that would be: 16cis180. This is because 16cos180 is -16, and 16 i sin180 is 0i. After converting it, we take the fourth root of 16 and divide 180 by 4 (a la DeMoirve's Theorem.) We are left with 2 cis 45. However, since the problem is asking for four answers, we need 3 more. So, we divide 360 by the number of roots it's asking for (in this case, that is four.) So, we get 90. Then, we add 90 to 45 3 times so we get 2 cis 45, 2 cis 135, 2 cis 180, and 2 cis 225. Finally, we must convert this to rectangular coordinates. So, for 2 cis 45, we'd calculate 2 cos 45 & 2 i sin 45 to get 2+ sqrt 2i. Repeat for the other 3 roots. And YAY you have an answer! Hope this helped.
$\endgroup$ $\begingroup$So let's use the formula: z=(fourth root of -16) times (cos x + i(sin (x))) let k = 0, solution 1= 2(radical i) (cos (0 + 2(180)(0)/(4) + i(sin (0+2(180)(0)/(4))= Solution 1: 2 (radical i) let k = 1, solution 2= 2 (radical i) (cos (0 +2(180)(1)/(4) + i(sin(0+2(180)(1)/(4)) Solution 2: 2(i)(radical i) let k = 2, solution 3= 2 (radical i) (cos (0+2(180)(2)/(4) + i(sin (0+2(180)(2)/(4)) Solution 3: -2(radical i) let k = 3, solution 4= 2 (radical i) (cos (0+2(180)(3)/(4) + i(sin (0+2(180)(3)/(4)) Solution 4= -2(i)(radical i) So the solutions are imaginary pairs (conjugates) and they are: 2 (radical i), 2i(radical i), -2(radical i), and -2i(radical i). If you try them, they will all work. I solved it using De Moivre's Theorem.
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