Find all the fourth roots of i without using De Moivre
I need to find all the fourth roots of i without using De Moivre's theorem and this is what I have so far but I'm stuck. I want to know if this is a good approach or there's a better way of doing it. Thanks.
$$\sqrt[4]{i} = \sqrt[4]{a+bi} $$ $\iff i = (a+bi)^4 $
$\iff a^4+4a^3b\cdot i+6a^2b^2\cdot i^2 + 4ab^3\cdot i^3 + b^4\cdot i^4$
$\iff (a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)\cdot i$
Thus $(a^4-6a^2b^2+b^4)$ must be equal to $0$ and $(4a^3b-4ab^3)$ must be equal to one so it satisfies the equation, it would be very helpful if you could help me.
$\endgroup$ 24 Answers
$\begingroup$Hint:
As $b\ne0,$ divide both sides of the first relationships by $b^4$ to find
$$(a^2/b^2)^2-6(a^2/b^2)+1=0$$
Solve for $a^2/b^2$
Replace the value of $b$ in terms of $a$ in the second relation.
$\endgroup$ $\begingroup$Start by taking the square roots of $i$: these will be $s+it$ where $s^2 - t^2 = 0$ (so $s = \pm t$) and $2 s t = 1$, and we get $\pm (1/\sqrt{2} + i/\sqrt{2})$. Then take the square roots of those.
$\endgroup$ $\begingroup$This is a fine approach. You now have two equations with two unknowns, so you can solve it algebraically.
$\endgroup$ 1 $\begingroup$To compute four roots of $ z^4-1 =0. $ When factored $ (z-1)(z+1)(z^2+1)=0 $
the roots are
$$ z =+1,-1, i, -i, $$
equi-spaced along $ (x,y) $ axes in the complex plane.
To compute four roots of $ z^4+1 =0 $ when factored as $ (z-\alpha )(z-\beta)(z-\gamma )(z-\delta)$
they are
$$[ \frac{ (1+ i)}{\sqrt2} \, , \frac{ (-1+ i)}{\sqrt2} \, , \frac{ (-1-i)}{\sqrt2} \, , \frac{ (1- i)}{\sqrt2} ]\,; $$.
Length of radius vector is unity in them all.
If $ (c,s) = (\cos u, \sin u) $ then the four roots can be also placed with a phase shift $u$ with respect to the first case:
$$ ( c,is) , (-s,ic),(-c,-is) (s,-ic ) $$
which roots come from the originating complex equation
$$(\cos 2u + i \sin 2u -z^2) \,( \cos u +i( z+ \sin u ))^2 =0 $$
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