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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.

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    Where is this claim on Wikipedia?2012-05-04
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    What happens to your result when $y$ is negative?2012-05-04
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    @Chris: I added the relevant link.2012-05-04
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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 $y$**."2012-05-04
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    @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*.2012-05-04
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    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.2012-05-04

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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.