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The results I get not the same as in the book, basically I need to get only this and the $x$'s of course: $$y = \frac{\sqrt{2}}{2}$$ So this is what I did: $$\sqrt{i} = x+yi$$ $$i = x^2 +2xyi -y^2$$

$$\begin{cases} x^2-y^2 = 0 \\ 2xy = 1 \end{cases}$$

$$xy=0.5$$

$$x=\frac{0.5}{y}$$

$$(1)\space\space\space\space(\frac{0.5}{y})^2-y^2=0$$ $$0.25-y^4=0$$ $$y^4=0.25$$ $$y=\pm \sqrt[4]{0.25}$$ $$y=\pm \frac{\sqrt{2}}{2}$$

What is WRONG?

Michael Hardy
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LiziPizi
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  • As far as I can see, nothing. There are two square roots of i. – Zain Patel Sep 10 '15 at 12:51
  • @ZainPatel General question, does square roots always have two solutions? any special cases? – LiziPizi Sep 10 '15 at 12:53
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    This might be helpful: http://math.stackexchange.com/q/3315/26188 – Thomas Sep 10 '15 at 12:56
  • There are always two square roots to a number in the complex plane (except for $0$.) We sometimes pick arbitrarily the square root $x+yi$ with $x>0$, or, if $x=0,$ with $y\geq 0$. – Thomas Andrews Sep 10 '15 at 13:11
  • @LiziPizi The only special case is 0, where the two square roots are equal. – MJD Sep 10 '15 at 13:12
  • "Solving $\sqrt i$" is not a correct usage; one could refer to "Evaluating $\sqrt i$" or "Finding $\sqrt i$". Or even to "Solving the problem of finding $\sqrt i$" or maybe even "Solving the problem of $\sqrt i$" (if you really want the word "solve" in here somewhere). One solves problems; one solves equations. One evaluates expressions. The expression $\text{“}\sqrt i\ \text{"}$ is not an equation. ${}\qquad{}$ – Michael Hardy Sep 10 '15 at 13:56

3 Answers3

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What you have shown is that there are two different complex numbers whose square is $i$. I am guessing that your confusion is that you think of the square root as a function. That is, for each input, you should only get one output.

For real a real number $a >0$ you also have two numbers whose square is $a$. We choose the positive of these are the square root of $a$.

For complex numbers things are a bit more tricky because you again have two different solutions to an equation $z^2 = a$. For example $i^2 = 1$ and $(-i)^2 = 1$. So how do you pick which one should be $\sqrt{-1}$?

For this question, please see this question/answer: How do I get the square root of a complex number?

The basic answer is that square roots of complex numbers aren't in general defined.

Well what about $\sqrt{-1}$ then? Here we usually defined $\sqrt{-a}$ (for $a >0$) to be $\sqrt{a}i$. It is just a matter of definition.

Community
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Thomas
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It would be nice to pick a rule which makes the square root function "smooth" in some sense.

So that $\sqrt{1.01 i} \approx \sqrt{i}$.

However this is usually difficult to accomplish over the whole set of numbers. Therefore one often needs to make a branch cut, cutting the set of numbers along some curve so that the function becomes discontinous there. It is often possible to choose this curve in many different ways and not always easy to say which is the best.

mathreadler
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You have $y = \pm \dfrac{\sqrt 2} 2$.

If $y= \dfrac{\sqrt 2} 2$ then $x=\dfrac{0.5} y = \dfrac{0.5}{\sqrt 2/2}$, and this simplifies to $x=\dfrac{\sqrt 2} 2$.

So $\dfrac{\sqrt 2} 2 + i \dfrac{\sqrt 2} 2$ is a complex number whose square is $i$.

If you multiply that by $-1$ you get the other complex number whose square is $i$. That is the one that comes from $y=\dfrac{-\sqrt 2} 2\vphantom{\dfrac{\displaystyle\int}\int}$.

One can show that there cannot be more than two solutions: Recall from algebra that if $a$ is a solution to $x^2 -i=0$ then $x-a$ is a factor of $x^2-i$, so you get $$ x^2 - i = (x-a)(\cdots\cdots). $$ The other factor must be a first-degree polynomial, so it can't have more than one root.

Michael Hardy
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