Can you enter the rabbit hole recursively?

If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?

  • 3
    $\begingroup$ It is not a nice practice to write $\sqrt a$ whenever $a$ is either a negative real number, or a complex number. $\endgroup$ – Pedro Tamaroff Mar 26 '13 at 1:36
  • $\begingroup$ $i$ as a factor represents a rotation by $90°$. Then $\sqrt{-i}$ is a rotation by $-45°$. From analytical geometry, $(x,y)\to\frac1{\sqrt2}(x+y,x+-y)$, which corresponds to the product $(x+iy)(\frac1{\sqrt2}-\frac i{\sqrt2})$. $\endgroup$ – Yves Daoust Feb 2 '17 at 21:01

I'll add an answer simply because I think it would be helpful or interesting to the original poster to eliminate the exponential. Drawing from either answer by Marvis or Andyk, we have $$e^{i3\pi/4} = \frac{\sqrt{2}}{2}(-1+i), ~ e^{i7\pi/4} = \frac{\sqrt{2}}{2}(1-i) \quad\Longrightarrow\quad \sqrt{-i} = \pm\tfrac{\sqrt{2}}{2}(1-i).$$ Note that unlike with real numbers, you cannot select a principal square root; that is, neither choice is "better" than the other. Working backwards to prove this is correct: $$\left(\pm\frac{\sqrt{2}}{2}(1-i)\right)^2 = \frac{1}{2}(1-i)^2= (1-2i+i^2)/2 = -2i/2 = -i.$$ EDITED TO ADD: In a comment above, Peter Tamaroff says "It is not a nice practice to write $\sqrt{a}$ whenever a is either a negative real number, or a complex number." This is because $\sqrt{\cdot}$ is used to refer to the principal square root of a number, which for positive numbers refers to the nonnegative root. That is, if $x^2=9$, then we know that $x$ could be $+3$ or $-3$; but the principal square root is $\sqrt{9}=3$. There is no corresponding convention to define the principal square root of a negative or complex number---though to be frank, for negative numbers I've seen $\sqrt{-a} \triangleq \sqrt{a}\cdot i$ adopted in more casual discussion.


We have $$i = e^{2k \pi i+ i \pi/2}$$ Hence, $$-i = e^{2k \pi i - i \pi/2} \implies (-i)^{1/2} = \left(e^{2k \pi i - i \pi/2}\right)^{1/2} = e^{k \pi i - i \pi/4} = e^{i 3 \pi/4}, e^{i 7 \pi/4}$$ We have two possible values, just like we have two possible values for $x^2 = -1$.

  • $\begingroup$ Thank you for the explanation; this is a new area of mathematics for me. I really thought that I had found an infinite recursion. The fact that this sews everything together has left me happy. $\endgroup$ – Dave Kirkby Mar 26 '13 at 1:50
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    $\begingroup$ @DaveKirkby: Dear Dave, You may want to learn about the fundamental theorem of algebra. This guarantees that there is no rabbit hole: the problem of solving algebraic equations is solved once and for all by the complex numbers. Regards, $\endgroup$ – Matt E Mar 26 '13 at 2:16

We have $i^2=-1$, thus $\sqrt{-i}=\sqrt{i^3}=i^{3/2}$

Also $i=(-1)^{1/2}$, so $i^{3/2}=\pm(-1)^{3/4}$. But $-(-1)^{3/4}=(-1)^{4/4}(-1)^{3/4}=(-1)^{7/4}$.

$$\sqrt{-i}=i^{3/2} = \left\{ \begin{array}{rl} (-1)^{3/4}\\ (-1)^{7/4} \end{array} \right.$$

And because $-1=e^{i\pi}$, you can write this as $e^{i3\pi/4}$ or $e^{i7\pi/4}$, exactly what Marvis has found.


Let us assume there exist a complex number $m$ such that $m * m = -i$, then $m$ will be the answer we are looking for $\sqrt{-i}$.

we can expand $m$ as $a + i.b$, then we have $(a + i.b)^2 = -i$; expanding we get

$a^2 - b^2 + i.2ab = -i$   $[i^2 = - 1] $; to make this true we must have

$a^2 - b^2 = 0$ --> (1)
$a * b = -1/2$ --> (2)

From (2) we have $b = -1/2a$ substituting in (1) we have
$a^2 - (-1/2a)^2 = 0$
rearranging we get the below
$a2 - (1/4a^2) = 0$
Multiplying both sides of equation with $a^2$ we get
$a^4 = 1/4$ and so $a=\pm1/\sqrt{2}$ we know $b=-1/2a$
So, when $a = 1/\sqrt{2}$ then $b = - 1/\sqrt{2}$
like wise when $a = -1/\sqrt{2}$ then $b = 1/\sqrt{2}$

Finally substituting the $a$ and $b$ values we get two roots for $\sqrt{-i}$

$$(-1/{\sqrt{2}}) + i.(1/\sqrt{2})$$ and $$(1/\sqrt{2}) - i.(1/\sqrt{2})$$
You'll see squaring any of the above solution you'll get $-i$


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