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I just want to clarify something here. Using elementary computation we can verify that for $x,y\in\mathbb{R}$

$$\sqrt{x+iy}=\pm\left(\sqrt{\frac{r+x}{2}}+i \sqrt{\frac{r-x}{2}}\right)$$

where $r=\sqrt{x^2+y^2}$.

However, in wikipedia the algebraic formula for the root is given by $$\sqrt{x+iy}=\sqrt{\frac{r+x}{2}}\pm i \sqrt{\frac{r-x}{2}}$$

By squaring $\sqrt{\frac{r+x}{2}}- i \sqrt{\frac{r-x}{2}}$ , We will get $x+iy=x-i\sqrt{y^2}$.

Wikipedia refers to the book Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Table , the book also giving the same formula as wikipedia's.

Is this something related to "Principal Value" or just a double typographic error?

I ask this because it is rare to see two same mistakes from two different sources, so I have a doubt.

Brian M. Scott
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Ajat Adriansyah
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    Where is this claim on Wikipedia? – Chris Eagle May 04 '12 at 07:33
  • What happens to your result when $y$ is negative? – Henry May 04 '12 at 07:35
  • @Chris: I added the relevant link. – Brian M. Scott May 04 '12 at 07:38
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    You apparently missed the stipulation stated on formula 3.7.27 in Abramowitz and Stegun: "$z^\frac12=\left(\frac12(r+x)\right)^\frac12\pm i \left(\frac12(r-x)\right)^\frac12=u\pm iv$ where $2uv=y$ and where the ambiguous sign is taken to be the same as the sign of $y$." – J. M. ain't a mathematician May 04 '12 at 07:38
  • @J.M.: And in the Wikipedia article, which says where the sign of the imaginary part of the root is taken to be same as the sign of the imaginary part of the original number. – Brian M. Scott May 04 '12 at 07:41
  • ooopps now I get it.. and the answer below (by robjohn) is a nice derivation.. to justify the square, note that $x+iy=x-\sqrt{y^2}i = x-|y|i$, now for $y<0$ we have $|y|=-y$, and the identity (with $sign(y)$) is confirmed. – Ajat Adriansyah May 04 '12 at 08:34

1 Answers1

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Since $r\ge|x|$, $$ \left(\sqrt{\dfrac{r+x}{2}}\right)^2=\dfrac{r+x}{2} \text{ and } \left(\sqrt{\dfrac{r-x}{2}}\right)^2=\dfrac{r-x}{2} $$ However, since $y$ may be positive or negative, $$ \sqrt{\dfrac{r+x}{2}}\sqrt{\dfrac{r-x}{2}}=\dfrac{\sqrt{r^2-x^2}}{2}=\dfrac{|y|}{2} $$ Thus, $$ \left(\sqrt{\dfrac{r+x}{2}}+i\sqrt{\dfrac{r-x}{2}}\right)^2=x+i|y| $$ If $y>0$, $$ \left(\sqrt{\dfrac{r+x}{2}}+i\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy $$ However, if $y<0$, $$ \left(\sqrt{\dfrac{r+x}{2}}-i\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy $$ So to be correct, we should incorporate $\newcommand{\sgn}{\operatorname{sgn}}\sgn(y)$: $$ \left(\sqrt{\dfrac{r+x}{2}}+i\sgn(y)\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy $$ Negate as necessary to get both solutions.

robjohn
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