Algebra simplifcation error with $\sqrt{-i}$

Question

I'm arriving at two different results when attempting to calculate $\sqrt{-i}$:

Method 1:

\begin{align} \sqrt{-i} = (-i)^{1/2} = \left( -\left(e^{i \pi /2}\right) \right)^{1/2} = (-1)^{1/2}\left( e^{i \pi/2} \right)^{1/2} = ie^{i \pi /4} = e^{i \pi /2}e^{i \pi /4} = e^{3\pi/4} \end{align}

Method 2:

\begin{align} \sqrt{-i} = \left( e^{- \pi /2} \right)^{1/2} = e^{- i \pi /4} \end{align}

What gives? I trust that Method 2 is the correct (and much more straightforward) simplification but I cannot find an error in my line of reasoning in Method 1. I think the issue in Method 1 comes from the fact that $-i = - (e^{i \pi /2})$ and $-i = e ^{- \pi /2}$

Answer 1

Martin R (and now Sofia) answer your question. But here's an even easier example for how this fails. Beware, wrong calculation incoming:

$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i^2 = -1$

This is clearly false. The reason is that, while one can still define a squareroot function for things that are not positive real numbers, this new function will no longer be multiplicative, and therefore, you cannot write $\sqrt{ab} = \sqrt{a}\sqrt{b}$.

PS: informally, the reason is basically the same as the one explained in Martin R's linked post. The complex squareroot is not a priori single-valued, and one hence has to make a choice of branch cut. If we're in situations where our 'total angle' (so the module of $ab$) is small enough, then this should not present an issue. But in cases where it is too big, then we end up mixing information from two different branch cuts, and hence make mistakes.

Answer 2

As was pointed out by Martin R, every nonzero complex number has two complex square roots. Both the answers found are correct, although one needs to be careful when aplying the property $(a^b)^c = a^{bc}$ to complex powers (see For which complex $a,\,b,\,c$ does $(a^b)^c=a^{bc}$ hold?).

To see that both answers are indeed correct, note that

\begin{align*} (e^{i3\pi/4})^2 &= \left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)^2 \\ &= \frac{1}{2} - i - \frac{1}{2} \\ &= -i, \end{align*}

and

\begin{align*} (e^{-i\pi/4})^2 &= \left(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right)^2 \\ &= \frac{1}{2} - i - \frac{1}{2} \\ &= -i. \end{align*}

Answer 3

Given any $~z \in \Bbb{C_{\neq (0 + i0)}},~$ the standard procedure that I was taught, from "An Introduction To Complex Function Theory" (Bruce Palka), to computing the $~n~$ distinct roots of $~z,~$ which are comprised of the $~n~$ distinct solutions of the equation $~w^n = z ~: ~w \in \Bbb{C}, ~n \in \Bbb{Z_{\geq 2}}$ is as follows:

• Identify $~\theta \in (-\pi, \pi],~$ a half open interval of width $~2\pi,~$ and identify $~r \in \Bbb{R^+} ~$ such that $~z = re^{i\theta}.$
  
  An alternative half-open interval that suggests itself for $~\theta~$ is $~[0,2\pi).$
  
  These two (arbitrary) alternatives for the range of $~\theta~$ suggest themselves for no other reason that they will prove convenient for the remainder of the procedure.

• For $~k \in \{0,1,\cdots,n-1\},~$ set $~\xi_k~$ as follows:
  $~\xi_0 = 1 + i(0).$
  $~\xi_k = e^{i2k\pi/n} ~: ~k \in \{1,\cdots,n-1\}.~$
  
  This implies that $~\{ ~\xi_0, ~\xi_1, ~\cdots, ~\xi_{n-1} ~\}~$ are the $~n~$ distinct solutions to $~z^n = 1 ~: ~z \in \Bbb{C}.$

• Identify $~\alpha = \theta/n.~$
  This implies that $~\alpha~$ is the smallest angle (in magnitude) such that $~e^{in\alpha} = [ ~e^{i\alpha} ~]^n = e^{i\theta}.~$ It is understood that if $~\theta \neq 0,~$ then $~\alpha \neq 0.~$

• So, one of the roots to the equation $~w^n = z~$ is $~r^{1/n} \times e^{i\alpha}.~$ Denote this root as $~w_1.~$ Here, it is intended that $~r^{1/n}~$ represent the unique element $~s \in \Bbb{R^+},~$ such that $~s^n = r.$

• Then, the $~n~$ distinct roots of $~w^n = z~$ are comprised of the $~n~$ distinct elements in
  $\{ ~(w_1 \times \xi_0), ~(w_1 \times \xi_1), ~\cdots, ~(w_1 \times \xi_{n-1}) ~\}.$

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For the present problem:

• $n = 2, ~\{ ~\xi_0, \xi_1 ~\} = \{ 1, -1\}. ~$

• $z = -i = e^{-i\pi/2} \implies r = 1, ~\theta = -\pi/2.~$
  Alternatively, $ ~\theta = 3\pi/2.~$

• So, $~\alpha = -\pi/4, ~$ or alternatively, $~\alpha = 3\pi/4.~$

• Then $~w_1 = e^{-i\pi/4}~$ or alternatively, $~w_1 = e^{i3\pi/4}.~$

• So, the $~2~$ solutions to $~w^2 = -i~$ are given by:
  $1 \times w_1 = 1 \times e^{-i\pi/4} = e^{-i\pi/4}.~$
  $-1 \times w_1 = -1 \times e^{-i\pi/4} = e^{-i5\pi/4}.~$
  
  Alternatively, the $~2~$ solutions to $~w^2 = -i~$ may be expressed as
  
  $1 \times w_1 = 1 \times e^{i3\pi/4} = e^{i3\pi/4}.~$
  $-1 \times w_1 = -1 \times e^{i3\pi/4} = e^{i7\pi/4}.~$

Answer 4

Shortcut method:- $$\sqrt -i =\sqrt{\frac{-2i}{2}}=\frac{\sqrt {1-2i+i^2}}{\sqrt 2}=\frac{1-i}{\sqrt 2}=e^{-i\frac π4}$$

This is your only answer. Fyi, the other root, $-\sqrt{-i}$ is nothing but this value rotated $90°$, or multiplied by $e^{\frac π2}$.

Also, if $v$ is the square root of $u$, with $u$ being a complex number then

$$(x_v+iy_v)^2=x_u+iy_u \implies x_v^2-y_v^2+2x_vy_vi=x_u+iy_u$$

Thus $(x_v^2-y_v^2)=x_u,2x_vy_v=y_u$

We get $(x_v^2+y_v^2)=\sqrt{x_u^2+y_u^2}$ Allowing us to get $$x_v=±\sqrt\frac{|u|+\Re(u)}{2},y_v=±\sqrt\frac{|u|-\Re(u)}{2}$$ Thus $$v=±\bigg(\sqrt\frac{|u|+\Re(u)}{2}+\sqrt\frac{|u|-\Re(u)}{2}\bigg)\text{(if $y_u>0$)}$$ $$v=±\bigg(\sqrt\frac{|u|+\Re(u)}{2}-\sqrt\frac{|u|-\Re(u)}{2}\bigg)\text{(if $y_u<0$)}$$