#2**0 **

There will be two answers on the complex plane. I have not worked them out yet.

This is my initial attempt to make sense of it:

\(\sqrt{-i}=\sqrt{-1}*\sqrt i=i*\sqrt i = i^\frac{3}{2}=(\sqrt{-1})^\frac{3}{2}=((-1)^\frac{1}{2})^\frac{3}{2}=(-1)^\frac{3}{4}\)

But wolframalpha says that one of the answer presentations is \(-(-1)^\frac{3}{4}\)

http://www.wolframalpha.com/input/?i=sqrt(-i)

So even my inital attempt is not correct I do not where that negative sign out the front came from.

Melody
Oct 30, 2017

#4**+1 **

Simplify the following:

sqrt(-i)

Express -i as a square using -i = (1 - 2 i + i^2)/2, then look to factor.

-i = 1/2 - i - 1/2 = (1 - 2 i - 1)/2 = (1 - 2 i + i^2)/2 = (-i + 1)^2/2:

sqrt(1/2 (-i + 1)^2 )

For all n>=0, sqrt(m/n) = (sqrt(m))/(sqrt(n)).

sqrt(1/2 (1 - i)^2) = (sqrt((1 - i)^2))/(sqrt(2)):

(sqrt((1 - i)^2))/(sqrt(2))

For all complex z with Re(z)>0, sqrt(z^2) = z.

Cancel exponents. sqrt((1 - i)^2) = -i + 1:

(1 - i)/sqrt(2)

Multiply numerator and denominator of (-i + 1)/sqrt(2) by sqrt(2).

Rationalize the denominator. (-i + 1)/sqrt(2) = (-i + 1)/sqrt(2)×(sqrt(2))/(sqrt(2)) = ((-i + 1) sqrt(2))/(2):

**((-i + 1) sqrt(2))/(2) [Courtesy of Mathematica 11 - Home Edition]**

Guest Oct 31, 2017

#5**0 **

Thanks guest that makes sense.

I'll just write my version of what you have said in LaTex.

\(\begin{align} \sqrt{-i}&=\sqrt{\frac{1-2i+1}{2}}\\ &=\sqrt{\frac{1-2i-i^2}{2}}\\ &=\sqrt{\frac{(1-i)^2}{2}}\\ &=\frac{\sqrt{(1-i)^2}}{\sqrt 2}\\ &=\frac{1-i}{\sqrt 2}\\ &=\frac{\sqrt 2(1-i)}{2}\\ &=\frac{\sqrt 2}{2}-\frac{\sqrt 2\;i}{2}\\ \end{align}\)

Ok that is good although it is only one of the 2 answers

With the aid of a quick sketch I can see that the 2 roots are

\(\sqrt{-i}=\pm \left[\frac{\sqrt{ 2}}{2} (1-i)\right]\)

I still cannot see the error in my #2 answer though

Melody
Oct 31, 2017

#6**+1 **

Whenever you have to deal with roots or big powers of complex numbers, it's usually best, and is more often than not essential, to put the number into polar form and to make use of de'Moivre's theorem.

\(\displaystyle -i \equiv \cos(270+k.360)+i\sin(270+k.360)\), k an integer, (or zero), angles in degrees,

so

\(\displaystyle \sqrt{-i}=\{\cos(270+k.360)+i\sin(270+k.360)\}^{1/2}\\=\cos(135+k.180)+i\sin(135+k.180)\)

\(\displaystyle k=0: \; =\cos(135)+i\sin(135)= -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}},\)

\(\displaystyle k=1:\;=\cos(315)+i\sin(315)= \frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\).

Other integer values for k simply repeat these, so just two possibles.

Tiggsy

Guest Oct 31, 2017

edited by
Guest
Oct 31, 2017

#7**0 **

Thanks Tiggsy

You lost me when you halved everything. I mean when you found the square root by halving 270 etc.

Why did you do that? Why does it work?

Melody
Oct 31, 2017

#8**+1 **

Hi Melody.

You need to read up on De Moivre's theorem.

The basic theorem says that one value of \(\displaystyle (\cos \theta+i\sin\theta)^{n} \text{ is }\cos n\theta+i\sin n\theta. \)

That's easy to prove when n is an integer, just use induction.

The 'one value of ', (because the angle is multivalued), becomes important when n is a fraction.

The basic result above could be expressed as

\(\displaystyle \{\cos(\theta+2k\pi)+i\sin(\theta + 2k\pi)\}^{n}=\cos(n\theta+2kn\pi)+i\sin(n\theta+2kn\pi)\).

k is any integer, and if n is also an integer, the 2kpi bit isn't needed, but if n is a fraction, (the theorem still holds if this the case), you now have fractional parts of 2pi, and therefore multiple values for the roots.

For square roots the numbers are 180 degees apart so there are just two possibles, cube roots are 120 degrees apart so there are three possibles, and so on.

Tiggsy

Guest Oct 31, 2017