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The complex number $z$ is given by $z = -1 + (4 sqrt{3})i$

The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.

So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.

How can I find $a$ and $b$ without a calculator?

The second equation can be written $ab=2sqrt{3}$ which gives $b = frac{2sqrt{3}}{a}$. If we substitute back into the first equation we get $a^2 - frac{12}{a^2} = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.

Observe that $|z| = sqrt{(-1)^2+(4sqrt3)^2} = sqrt{49} = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.

Note that$$z=7left(-frac17+frac{4sqrt3}7iright).tag1$$Now, since $left(-frac17right)^2+left(frac{4sqrt3}7right)^2=1$, the expression $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the square roots of $z$.

One way is to write $z=r^2 e^{2theta}$ and roots will be $re^theta$ and $re^{pi -theta}$.

From $z =-1+4sqrt{3}i$, we obtain $r=7$ and $tan{2theta} = frac{2tantheta}{1-tan^2theta} = -4sqrt{3}$. Second expression gives you a quadratic equation, $2sqrt{3}tan^2theta -2sqrt{3} + tantheta =0$.

Roots of the above quadratic equation are $tantheta= sqrt{3}/2,-2/sqrt{3}$ which form $tantheta$ and $tan(pi-theta)$.

Hence, square roots of $z$ are $(1+sqrt{3}/2i)frac{7}{sqrt{1+3/4}} = sqrt{7}(2+sqrt{3}i)$ and $(1-2/sqrt{3}i)frac{7}{sqrt{1+4/3}} = sqrt{7}(sqrt{3}-2i)$.

Let$$
z^2=(x+yi)^2=−1+4sqrt3i,
$$
i.e.$$
(x^2-y^2)+2xyi=−1+4sqrt3i.
$$
Compare real parts and imaginary parts, $$
begin{cases}
x^2 - y^2 = -1&qquadqquad(1)\
2xy = 4sqrt3&qquadqquad(2)
end{cases}
$$
Now, consider the modulus: $|z|^2 =|z^2|$, then
$$x^2 + y^2 = sqrt{smash[b]{(-1)^2+(4sqrt3)^2}} = 7tag3$$
Solving $(1)$ and $(3)$, we get $x^2 = 3Rightarrow x = pmsqrt3$ and $y^2 = 4Rightarrow y = pm2$.

From $(2)$, $x$ and $y$ are of same sign,
$$begin{cases}
x = sqrt3\
y = 2
end{cases}text{ or }
begin{cases}
x = -sqrt3\
y = -2
end{cases}
$$
then$$z = pm(sqrt3 + 2i).$$

That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
begin{align}
& left|-1 + i4sqrt 3right| = sqrt{(-1)^2 + (4sqrt 3)^2 } = 7. \[10pt]
text{Therefore } & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
text{Therefore } & pmsqrt{-1+i4sqrt 3} = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
end{align}

Notice that
$$
sin varphi = frac{4sqrt 3} 7 quad text{and} quad cosvarphi = frac{-1} 7
$$
and recall that
begin{align}
tanfracvarphi 2 & = frac{sinvarphi}{1+cosvarphi} \[12pt]
text{so we have }tanfracvarphi 2 & = frac{4sqrt 3}{7-1} = frac 2 {sqrt 3}. \[10pt]
text{Therefore } sinfracvarphi2 & = frac 2 {sqrt 7} quad text{and} quad cosfracvarphi2 = frac{sqrt3}{sqrt7}.
end{align}
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$