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A question in my text book states as follows:

Find all possible values of $\sqrt i + \sqrt{-i}$.

Firsly I found the value of $\sqrt i$ which came out to be $\pm(1/\sqrt 2 + i(1/\sqrt 2))$. Then I wrote $\sqrt{-i}$ as $\sqrt{-1} \sqrt{i} = i\sqrt{i}$. So the given expression became $\sqrt{i} (1+i)$. In this I put the two values of $\sqrt{i}$ and hence got $2$ answers. But the textbook states $4$ answers for this question. Can you please let me know in which step I am wrong.

Gary
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Madhav Mittal
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    Just a remark: in general, $\sqrt{ab}\neq \sqrt{a}\sqrt{b}$ for complex $a$ and $b$. – Gary Jan 22 '24 at 08:19

3 Answers3

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Your textbook is wrong:

For real numbers ($a \in \mathbb{R}$), the expression $\sqrt{a}$ makes sense:

$b = \sqrt{a} \iff b \ge 0 \text{ and } b^2=a$

For complex numbers, this is impossible because the expression $c \ge 0$ does not make sense, hence the term $\sqrt{a}$ does not make sense for complex numbers, hence the term $\sqrt{i}$ does not make sense.

Yet, the expression gets used sometimes, and in fact it means that the person is looking for $x$, for which: $x^2=i$.

As you know, the equation $x^2=i$ has two solutions for $x$, and also the equation $x^2=-i$ has two solutions, so there should be four solutions for $x^2=i \text{ or } x^2=-i$, and honestly I believe you already found all solutions:

For $x^2=i$, you have found two solutions:

  1. $x_1=\frac{\sqrt{2}}{2}(1+i)$
  2. $x_2=-\frac{\sqrt{2}}{2}(1+i)$

You said yourself that, in order to find the solutions of $x^2=-i$, you just multiply the solutions by $i$, so let's to that:

  1. $x_3 = i \cdot x_1 = i \cdot \frac{\sqrt{2}}{2}(1+i) = \frac{\sqrt{2}}{2}(i-1)$
  2. $x_4 = i \cdot x_2 = i \cdot (-\frac{\sqrt{2}}{2}(1+i)) = -\frac{\sqrt{2}}{2}(i-1)$
Dominique
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Call $x=\sqrt i-\sqrt{-i}$. Then $$ \begin{array}{l} x-\sqrt{i}=\sqrt{-i}\\ x^2-2\sqrt{i}x+i=-i\\ x^2+2i=2\sqrt{i}x\\ x^4+4ix^2-4=4ix^2, \end{array} $$ thus $x$ is any of the roots of $x^4-4$.

Andrea Mori
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$\sqrt{-1}$ can be $\pm i$ and therefore you get 4 answers corresponding to all the combinations of $\pm$ in your two terms

Sorin Tirc
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    How is √(-1) equal to (-i)? – Madhav Mittal Jan 22 '24 at 08:37
  • @MadhavMittal Because $(-i)^2 = -1$, just as $i^2 = -1$, so they are both square roots of $-1$. (This confusion of yours is pretty common, and it's one big reason I generally try to discourage people from using square root symbols on numbers that aren't non-negative reals.) – Arthur Jan 22 '24 at 08:51