Write the exponential form of $\sqrt{i}$.

Question

Question: Write the exponential form of $\sqrt{i}$.

My approach: Let $z=x+iy$ be such that $e^z=\sqrt{i}$. Thus we have $e^{2z}=i$. Hence we have $e^{2x+2iy}=i.$ This also implies that $|e^{2z}|=e^{2x}=1\implies 2x=\ln 1=0\implies x=0.$

Thus we have $e^{2x}\cos (2y)=\cos(2y)=0$ and $e^{2x}\sin (2y)=\sin(2y)=1.$ Thus $2y=\frac{\pi}{2}\pm 2n\pi, n\in\mathbb{W}$, which in turn implies that, $y=\frac{\pi}{4}\pm n\pi.$

Thus we have $\Re(z)=0$ and $\Im(z)=\frac{\pi}{4}\pm n\pi$.

Now observe that, $$e^z=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i,\text{ when $n$ is even, and}\\ e^z=-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}i,\text{ when $n$ is odd}.$$

Also observe that indeed $e^{2z}=i,$ in both the cases. Hence, we can conclude that $$\sqrt i=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i, -\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}i.$$ Thus in exponential form, $$\sqrt{i}=e^{\frac{\pi i}{4}},e^{\frac{5\pi i}{4}}.$$

Is this solution correct and is there a better way to solve this problem?

Answer 1

Not quite: you want $\pm e^{\pi i/4}$, so each of your exponents needs a factor of $i$.

Answer 2

$e^{i\pi}=-1$

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

$e^{\frac{i\pi}{4}}=\sqrt{i}$

Now think about writing a general expression in terms of $n \in \mathbb{Z}$.

$e^{\frac{i\pi}{4}+2\pi n}=\sqrt{i}$

Answer 3

You mustn't use radicals when deeling with complex numbers.

If $ n $ is a positive integer than the function $ f_{n} : x\mapsto\sqrt[n]{x} $ is well-defined only on $ \mathbb{R}_{+}=\left[0,+\infty\right) \cdot $

To answer your question, $ \mathrm{e}^{\mathrm{i}\frac{\pi}{4}} $, and $ \mathrm{e}^{-\mathrm{i}\frac{3\pi}{4}} $ are both square roots of $ \mathrm{i} \cdot $

Answer 4

I make this an answer to healynr, since comment section is starting to be long:

In $\mathbb C$ the fundamental theorem of algebra states that $x^2=i$ or $x^2-i=0$ has exactly $2$ roots since the degree of the polynomial is $2$.

While in $\mathbb R$, the quantity $\sqrt{x}$ designate specifically the positive solution, in $\mathbb C$ you are right that $\sqrt{z}$ can designate the principal value (i.e. when $k=0$ see later), but we generally prefer giving all the complex solutions since many reciprocal "functions" get multivalued in $\mathbb C$.

Even though there are still only two solutions :

$i=e^{\frac{i\pi}2+2ik\pi}\implies \sqrt{i}=e^{\frac{i\pi}4+ik\pi}$

This gives the principal value $\quad z_0=e^{\frac{i\pi}4}=\frac{\sqrt{2}}2+i\frac{\sqrt{2}}{2}\quad$ for $k=0$

And a second value $\quad z_1=e^{i\frac{\pi}4+i\pi}=e^{i\frac{5\pi}4}=-z_0\quad$ for $k=1$

Note that subsequent values are cylcing, $z_2=z_0$, $z_3=z_1$, etc.

In fact in your expression you have ommitted an $i$ thus $\quad e^{i\frac{\pi}4+i2n\pi}=z_0\times \underbrace{e^{i2n\pi}}_{=1}=z_0$

Similarly for $z^d=e^{i\theta}$ the roots will be given by $\quad e^{i\frac \theta d+i\frac{2n\pi}{d}}$ and there not an infinity when $n$ varies because it cycles when $n$ becomes greater than $d$.

The cases where there is an infinity of solutions is for instance for $i^i$, now you will get a real exponential depending on $n$ and it does not cycle.

Answer 5

The exponential form of $i$ is $e^{i\pi/2+i2k\pi}$. So that of $\sqrt i$ is $e^{i\pi/4+ik\pi}$. There are two distinct values,

$$e^{i\pi/4},e^{i5\pi/4}.$$