Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Where is this claim on Wikipedia? –  Chris Eagle May 4 '12 at 7:33
    
What happens to your result when $y$ is negative? –  Henry May 4 '12 at 7:35
    
@Chris: I added the relevant link. –  Brian M. Scott May 4 '12 at 7:38
2  
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. May 4 '12 at 7:38
    
@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 4 '12 at 7:41

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.