2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Where is this claim on Wikipedia? $\endgroup$ – Chris Eagle May 4 '12 at 7:33
  • $\begingroup$ What happens to your result when $y$ is negative? $\endgroup$ – Henry May 4 '12 at 7:35
  • $\begingroup$ @Chris: I added the relevant link. $\endgroup$ – Brian M. Scott May 4 '12 at 7:38
  • 2
    $\begingroup$ 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 $y$." $\endgroup$ – J. M. is a poor mathematician May 4 '12 at 7:38
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott May 4 '12 at 7:41
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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