Converting $\sqrt{\sqrt{i}}$ into the form of $a+bi$ with a system of equations

Question

Disclaimer: I am in tenth grade and this is my first time asking a question here, so please excuse my lack of knowledge and any mistakes of mine. I'm just really interested with this problem.

So, I watched a YouTube video by blackpenredpen about converting the $\sqrt{i}$ into the form of $a+bi$. So, I wanted to see whether I could try the next logical step of that: turning the $\sqrt{\sqrt{i}}$ into the form $a+bi$.

So far, I have done the following:

$\sqrt{i}=a^2+2abi-b^2$
$0+1i=a^4+4a^3bi-6a^2b^2-4ab^3i+b^4$
$0+1i=(a^4+b^4-6a^2b^2)+(4a^3b-4ab^3)i$

Which you could put into this system of equations:

$\begin{cases} a^4+b^4-6a^2b^2=0\\ 4a^3b-4ab^3=1 \end{cases}$

Then, I tried to solve this system by letting $b=mx$, where m is some constant. But, that didn't work. So, my question is the following: how can I solve the above system?

It'd be great if you could keep it as simple as possible, as I am only a tenth grader. Thanks!

Answer 1

The equations are correct, but as you noted it is (probably) hard to solve the system of equations. Therefore, I suggest another method, which I believe you will understand as a tenth grader :)

So as you wrote, we have $a^2 - b^2 + 2abi = \sqrt{i}$. Instead of expanding further, we try to compute $\sqrt{i}$ first.

Using the same method, we let $\sqrt{i} = x + yi$, so we get $i = (x^2 - y^2) + 2xyi$, which gives

$$\begin{cases} x^2 - y^2 &= 0 \\ 2xy &= 1 \end{cases}$$

So we have $x = \pm y$, which gives $z = x + yi = \pm\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right) = \pm\frac{1}{\sqrt{2}}(1 + i)$. Notice that we have two solutions! This is because we are solving the quadratic equation $z^2 = i$, so there will be two complex roots here (and they happen to be distinct).

Now, we can substitute this into our $a$-$b$ equation, which gives $a^2 - b^2 + 2abi = \pm\frac{1}{\sqrt{2}}(1 + i)$. I will leave the rest for you, but you should get $4$ different roots, one of will which be

$$z = \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2}i$$

---

In fact, let's generalise this and derive a formula for finding square roots of any complex number $a + bi$. That is, we want to solve

$$z^2 = a + bi$$

Let $z = x + yi$. Then,

$$\begin{align*} (x + yi)^2 &= a + bi \\ (x^2 - y^2) + 2xyi &= a + bi \\ \begin{cases} x^2 - y^2 &= a \\ 2xy &= b \end{cases} \\ x &= \frac{b}{2y} \\ \left(\frac{b}{2y}\right)^2 - y^2 &= a \\ b^2 - 4y^4 &= ay^2 \\ 4y^4 + ay^2 - b^2 &= 0 \end{align*}$$

You can solve this by doing a substitution of $u = y^2$, which gives

$$\begin{align*} 4u^2 + au - b^2 &= 0 \\ u &= \frac{-a \pm \sqrt{a^2 + 16b^2}}{8} \\ y &= \sqrt{\frac{-a \pm \sqrt{a^2 + 16b^2}}{8}} \end{align*}$$

From this, you can also work out $x = \frac{b}{2y}$.

Hope this helps and keep up with the curiosity :)

Answer 2

Convert to polar form:

$$i^{1/4}=(e^{2\pi ik})^{1/4}(e^{i\pi/2})^{1/4}=e^{\pi ik/2}e^{i\pi/8},k=0,1,2,3\\ =e^{\pi ik/2}(\cos(\pi/8)+i\sin (\pi/8)),k=0,1,2,3$$

giving solutions

$$\pm \left(\frac{\sqrt{2+\sqrt 2}}{2}+i\frac{\sqrt{2-\sqrt 2}}{2}\right),\pm \left(-\frac{\sqrt{2-\sqrt 2}}{2}+i\frac{\sqrt{2+\sqrt 2}}{2}\right).$$

Answer 3

It is easier to do it in two steps.

First, you compute the square roots of $i$. So, you're after real numbers $a$ and $b$ such that $(a+bi)^2=i$. And\begin{align}(a+bi)^2=i&\iff a^2-b^2+2abi=i\\&\iff\left\{\begin{array}{l}a^2-b^2=0\\2ab=1\end{array}\right.\\&\iff\left\{\begin{array}{l}(a-b)(a+b)=0\\ab=\frac12\end{array}\right.\\&\iff\left\{\begin{array}{l}a=\pm b\\ab=\frac12\end{array}\right.\end{align}and it is easy to see that the solutions are $\pm\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)$

Now, you find square roots of these two numbers. For instance, in order to find the square roots of $\frac1{\sqrt2}+\frac i{\sqrt2}$, you solve the system$$\left\{\begin{array}{l}a^2-b^2=\frac1{\sqrt2}\\2ab=\frac1{\sqrt2}.\end{array}\right.$$

Answer 4

I feel like other answers are not answering exactly what you are asking for. The first equation you have is of the form: $$x^4-6x^2+1 = 0$$ if you let $\dfrac{a}{b} = x.$ You can solve this if you know how to solve a quadratic. In fact, you get: $$x_{1,2}^2 = 3\pm2\sqrt{2} =(\sqrt{2}\pm 1)^2\implies x_{1,2,3,4} = \pm\sqrt{2}\pm 1.$$

Then you can use the second equation to get values for $a,b,$ although it would be easier if you remember that any root $a+bi$ must have modulus $1$ so $a^2+b^2 = 1.$ This would imply: $$b = \pm\dfrac{1}{\sqrt{1+x^2}} = \pm\dfrac{1}{\sqrt{4\pm 2\sqrt{2}}} = \pm\dfrac{\sqrt{2\mp\sqrt{2}}}{2}.$$