What is $\sqrt{i}$?

Solution 1:

Let $z=(a+bi)$ be a complex number which is a square root of $i$, that is $$i=z^2=(a^2-b^2)+2abi.$$ Equating real and imaginary parts we have, $$a^2-b^2=0, 2ab=1$$

The two real solutions to this pair of equations are $a={1 \over \sqrt{2}},b={1 \over \sqrt{2}}$ and $a=-{1 \over \sqrt{2}},b=-{1 \over \sqrt{2}}$. The two square roots of $i$ therefore are

$$\pm {1 \over \sqrt{2}} (1+i)$$

Solution 2:

Visually, the square root of a complex number (written in polar coordinates) $(\rho,\theta)$ is the number $(\sqrt\rho,\theta/2)$.

Solution 3:

$i^\frac1{2}=\left(e^{\pi i/2}\right)^\frac1{2}=e^{\pi i/4}$

$e^{\pi i/4}=\cos\left(\frac{\pi}{4}\right)+i \sin\left(\frac{\pi}{4}\right)$

or simplified,

$\frac{1+i}{\sqrt{2}}$

This is of course the "principal value"; the other value (thanks Matt E!) is the "negative square root", $-\frac{1+i}{\sqrt{2}}$ or in exponential form, $-e^{\pi i/4}=e^{-3\pi i/4}$

Solution 4:

As you can see, there are several ways to find the answer to your question, from jmoy's straightforward algebra calculation, to using Euler's formula $e^{r+i\theta} = e^r(\cos(\theta)+i \sin(\theta))$, to geometric interpretations of complex numbers.

One fact that hasn't been mentioned yet is that the complex numbers are algebraically closed. This means that every algebraic equation using complex numbers has all of its solutions in the complex numbers. So, the equation $x^2=i$ has both of its solutions in the complex numbers; and the equations $x^4 = -7-12i$ and $x^4+(\pi -8i)x^3+x-\sqrt{5}=0$ each have all four of their solutions in the complex numbers.

The real numbers are not algebraically closed: the equations $x^2=-1$ and $x^4+7x^2+\pi=0$ cannot be solved unless we use complex numbers.

This is one of the main reasons complex numbers are so important; they are the algebraic closure of the real numbers. You will never need "higher levels" of imaginary numbers or new mysterious square roots; numbers of the form $a+bi$ are all you need to find any root of real or complex polynomials.

Solution 5:

More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that

$$ z^n = z_0 \ , \qquad \qquad \qquad \qquad [1] $$

you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes

$$ \left( r e^{i \theta}\right)^n = r_0 e^{i\theta} \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0} \ . $$

Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:

$$ r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0} $$

since $r, r_0 \geq 0$.

As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:

$$ n\theta = \theta_0 + 2k\pi \qquad \Longleftrightarrow \qquad \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots $$

It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally

$$ \sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{\theta_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 $$

are all the complex $n$-th roots of $z_0$.

Examples

(1) For $n=2$, we obtain that every complex number has exactly two square roots:

$$ \begin{align} \sqrt{z_0} &= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1 \\\ &= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)} \ . \end{align} $$

For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain

$$ \sqrt{i} = \begin{cases} e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\\ e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \ . \end{cases} $$

Also, if $z_0 = -1 = e^{i\pi}$,

$$ \sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ . $$

(2) For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th roots of unity:

$$ \sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ . $$

For instance, if $n= 2$, we get

$$ \sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad e^{i\pi}= -1 $$

and for $n= 4$,

$$ \sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ , $$

that is,

$$ \sqrt[4]{1} = 1, i, -1 , -i \ . $$