\(\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\)
–
Pranav Arora
·
3 years, 10 months ago

Log in to reply

@Pranav Arora
–
remember taking root will give both positive and negative solutions
–
Manish Kansal
·
3 years, 10 months ago

Log in to reply

@Pranav Arora
–
I completely agree though you forgot to put in the \(i\) until the end: \(i=e^{\pi i/2}\) Voted up. :D
–
Joel Jablonski
·
3 years, 10 months ago

Actually, \(i\) has two square roots, just as any real number does. These are the two roots of \(z^2=e^{\pi i/2}\). One of the roots is \(e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) as noted previously. The other is \(e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\). Doubling the argument of both of these complex numbers and squaring the modulus indeed gives \(e^{\pi i/2}\).
–
Aaron Doman
·
3 years, 10 months ago

\(e^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\)
–
Ton De Moree
·
3 years, 10 months ago

Log in to reply

@Aaron Doman
–
There is some method with which we agree on which root we're talking about with radicals, making radicals ambiguous but fractional exponents not so.

Sadly, I don't know exactly how we do decide. Could anyone enlighten me?
–
Jonathan Wong
·
3 years, 10 months ago

Log in to reply

@Jonathan Wong
–
The "principal root" of x describes the positive root, so if you're given \( \sqrt{4}\) then it is asking for the positive answer which is \(2\) and not \(-2\). but when dealing with complex numbers it gets.. well, more complex. Neither root is more "important" than the other, so you don't tend to worry about it as much.
–
Michael Tong
·
3 years, 10 months ago

## Comments

Sort by:

TopNewest\(\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) – Pranav Arora · 3 years, 10 months ago

Log in to reply

– Manish Kansal · 3 years, 10 months ago

remember taking root will give both positive and negative solutionsLog in to reply

– Joel Jablonski · 3 years, 10 months ago

I completely agree though you forgot to put in the \(i\) until the end: \(i=e^{\pi i/2}\) Voted up. :DLog in to reply

Thanks! :) – Pranav Arora · 3 years, 10 months ago

Log in to reply

Actually, \(i\) has two square roots, just as any real number does. These are the two roots of \(z^2=e^{\pi i/2}\). One of the roots is \(e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) as noted previously. The other is \(e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\). Doubling the argument of both of these complex numbers and squaring the modulus indeed gives \(e^{\pi i/2}\). – Aaron Doman · 3 years, 10 months ago

Log in to reply

\(e^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\) – Ton De Moree · 3 years, 10 months ago

Log in to reply

Sadly, I don't know exactly how we do decide. Could anyone enlighten me? – Jonathan Wong · 3 years, 10 months ago

Log in to reply

– Michael Tong · 3 years, 10 months ago

The "principal root" of x describes the positive root, so if you're given \( \sqrt{4}\) then it is asking for the positive answer which is \(2\) and not \(-2\). but when dealing with complex numbers it gets.. well, more complex. Neither root is more "important" than the other, so you don't tend to worry about it as much.Log in to reply

– Jonathan Wong · 3 years, 10 months ago

and for higher roots? (e.g. \(\sqrt[5]{2}\))Log in to reply

let \(\sqrt{i}\) = \(x\)

we have \(x^{2}\) = \(i\)

converting i to polar form, we have

\(x^{2}\) = \(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}\)

using Demoivre's Theorem

we have \(x_{1} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} i\) and \(x_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i\) – Ed Mañalac · 3 years, 10 months ago

Log in to reply

\(\sqrt{i}\) is a complex number, so it can be written in the form \(a+bi\) where \(a\) and \(b\) are real numbers.

Squaring gives \((a^2-b^2)+2abi\). It follows that \(2ab=1\) and \(a^2=b^2\).

From the second we have that \(a=b\) or \(a=-b\).

Substituting in the first, we get \(a^2=\frac{1}{2}\) or \(a^2=-\frac{1}{2}\)

So \(a=\pm \frac{1}{2} \sqrt{2}\) or \(a=\pm \frac{1}{2} \sqrt{2}*i\).

and \(b=\pm \frac{1}{2} \sqrt{2}\) or \(b=\mp \frac{1}{2} \sqrt{2}*i\).

Using these in our original expression \(a+bi\), we get only two distinct solutions:

\( \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}*i \) and \(- \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2}*i \).

I think usually the first is chosen as THE root of \(i\), but only because it's closest to \((1,0)\). – Ton De Moree · 3 years, 10 months ago

Log in to reply