The Cube Root of $i$ ?

$i = e^{i\frac{\pi}{2}}$

$i^{1/3} = e^{i\frac{\pi}{6}}$

$i^{1/3} = \cos\frac{\pi}{6}+i\sin\frac{\pi}{6}$

$i^{1/3} = \frac{\sqrt{3}+i}{2}$

Hence,

$a+b = 5$

We can also use the n-th root formula

Let $z = r(\cos\theta + i\sin\theta)$ . For any positive integer $n$ , $z^{\frac{1}{n}} = r^{\frac{1}{n}}[\cos (\displaystyle\frac{\theta + 360^0k}{n}) + i\sin (\displaystyle\frac{\theta + 360^0k}{n})]$ , where $k= 0, 1, 2, ... n-1$ .

$i = \cos 90^0 + i\sin 90^0$ $i^{\frac{1}{3}} = (\cos 90^0 + i\sin 90^0 )^{\frac{1}{3}} = [\cos (\displaystyle\frac{90^0 + 360^0k}{3}) + i\sin (\displaystyle\frac{90^0 + 360^0k}{3})] = \cos(30^0 + 120^0k) + i \sin(30^0 + 120^0k)$

When $k = 0, 1, 2$ , the values of the new $\theta$ are $30^0, 150^0, 270^0$ . We only consider when $\theta = 30^0$ .

So we have

$\cos 30^0 + i \sin 30^0 = \displaystyle\frac{\sqrt{3}}{2} + i \displaystyle\frac{1}{2} = \displaystyle\frac{\sqrt{3} + 1}{2}$ . Thus, $a = 3, b=2$ then $a + b = 5$

Observe: $(\frac{\sqrt{a} + i}{b})^3 = i$ The bottom part is $b^3$ , lets look at the top part. $(\sqrt{a} + i)^3 = a\sqrt{a} - 3\sqrt{a} + 3ai - i$ Notice to cancel out the real part, put $a = 3$ Then the imaginary part is $9i - i = 8i$ . Notice $2^3 = 8$ , hence $8i / 8 = i$ . Hence $a = 3$ , $b = 2$ and $a+b = 5$

cube root of unity are 1 root of 3/2 * ioata and 1/2

we can write i = (cos pi/2 + isin pi/2 ) cos pi/2 = 0 and sin pi/2 =1 now its given (i)^1/3 so (i)^1/3 = (cos pi/2 + isin pi/2 )^1/3 . Now on applying De Moivre's Theorem we get (i)^1/3 = (cos pi/6+ isin pi/6 )= ( (3)^1/2 + i)/2 ... a=3 and b=2 so a+b =5