I'm trying to find the 4 fourth roots of 16, so far I have -2 and 2, but I have no idea how to find the other 2 complex numbers.
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$\begingroup$$x^4- 16 = 0$
$x^4 - 4^2 = 0$
$(x^2 - 4)(x^2 + 4) = (x^2 -4)(x^2 - (-4)) =(x^2 -4)(x^2 - (-1\cdot 4))0$
$(x^2 - 2^2)(x^2 - (i^2\cdot 2^2) = (x^2 -2^2)(x^2 - (2i)^2)=0$
$ (x-2)(x+2)(x-2i)(x+2i)=0$
so the four fourth roots are $2, -2, 2i, -2i$.
.....
Alternatively.
If $k^4 = 16$ and $k = a + bi$ where $a,b$ are real and $i^2 = -4$ we get.
$(a + bi)^4 = a^4 + 4a^3bi + 6a^2b^2i^2 + 4ab^3i^3 + b^4i^4=$
$a^4 + 4a^3bi + 6a^2b^2(-1) + 4ab^3(-1)i + b^4(-1)^2 = $
$(a^4- 6a^2b^2 +b^4) + (4a^3b-4ab^3)i = 16 = 16 + 0i$.
So $a^4 -6a^2b^2 +b^4 = 16$ and $4a^3b-4ab^3 = 0$.
Solve for $a,b$.
$a^3b - ab^3 = ab(a^2 -b^2)=0$ so
Either $ab =0$ or $a^2-b^2 =0$.
If $ab = 0$ then either $a = 0$ or $b=0$ and
And if $a^2 -b^2 =0$ then $a^2 = b^2$ and either $a = b$ or $a =-b$.
Case 1: $a = 0$ then
$b^4 = 16$ and $b=\pm 2$. So $a + bi = 0 \pm 2i$ so $2i$ and $-2i$ are two solutions.
Case 2: $b = 0$ then
$a^4 =16$ and $a =\pm 2$. So $a+bi = \pm 2 + 0i$ so $2$ and $-2$ are two more solutions.
Case 3: $a=b$ then
$a^4 - 6a^2b^2 + b^4= a^4 -6a^4 +a^2 = -2a^4 = 16$ so
$a^4 = -8$ . But $a$ is real and so there is no solution as no real number to the four power is negativve.
Case 4: $a = -b$ then
..... same thing. $a^4 -6a^2(-a)^2+(-a)^4 = -2a^4 = 16$ and there is no real $a$ where that is possible.
.....
Alternatively (although I suspect you have learned this yet. don't worry, you will.)
$16 = 16 + 0i$
$16(1 + 0i) =$
$16(\cos 2m\pi + i\sin 2m\pi ) := 16e^{2\pi ki}$
So if $k^4 = 16(\cos 2m\pi + i\sin 2m\pi )$ then
$k = 2(\cos \frac {2m\pi}2 +i\sin {2m\pi} )= 2e^{\frac {2m\pi}2 i}$.
(Because $(\cos \theta + i\sin\theta )^m = \cos m\theta + i\sin m\theta$..... do the trigonometry .....)
So if $m = 0,1,2,3$ then $k = 2(\cos 0 +i\sin 0), 2(\cos \frac \pi 2+i \sin\frac \pi 2), 2(\cos \pi + i\sin \pi), 2(\cos\frac {3\pi}2 + i\sin \frac {3\pi}2 ) =$
$2(1+0i), 2(0+1i), 2(-1+0i), 2(0-1i)=$
$2,2i, -2, -2i$.
$\endgroup$ $\begingroup$Trying solving this
$x^4 = 16$
$x^4-16 = 0$
Hint $:$
Just like $\sqrt[2]{n^2}= ±n$
$\sqrt[3]{n^3}= n$ or $(-1/2+(1/2)*\sqrt{-3})*n$ or $(-1/2-(1/2)*\sqrt{-3})*n$
$\sqrt[4]{n^4} = ±n$ or $±i*n$
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