Rewriting a complex number for use in an analytic computation

[CAF123]

Consider an equation, $$\tilde{x_0}
= \ln(X+ i\delta),$$ where X may be positive or negative and $0< \delta \ll 1$. Now, if $X>0$ this evaluates to $\ln(X)$ in some limiting prescription for $\delta \rightarrow 0$ while if $X<0$, we get $\ln(-X) + i \pi.$

Now, consider, $$\tilde{x_0} = 1-\sqrt{X+i \delta}.$$ How may I write this in the form $a + i \delta$, where a is real?

 

[anuttarasammyak]

For real number X
X=|X|sgn(X)=|X|\ e^{(1-sgn(X))\pi/2\ i}
so I think
\sqrt{X}=\sqrt{|X|}\ e^{(1-sgn(X))\pi/4\ i}
In short for X < 0
\sqrt{X}=i\sqrt{-X}

 

[Fred Wright]

You have $$

z_0=\left |Z\right |e^{i\theta} + i\delta $$
$$x_0 =\log(z_0) $$
$$\mathcal Re z_0= \left |Z\right |\cos(\theta) \nonumber $$
$$\mathcal I am z_0=\left |Z\right |\sin(\theta) + \delta \nonumber $$
$$\log(z_0)=\frac{1}{2}\log(\left |Z\right |^2+\delta^2)+i\arctan(\frac{\left |Z\right |\sin(\theta)+\delta}{\left |Z\right |\cos(\theta)})

$$
for $\log(-Z + i\delta)$ you have $\theta = \pi$ and$$\log(-Z + i\delta)=\frac{1}{2}\log(Z^2 + \delta^2)+i\arctan(\frac{\delta}{\left |Z\right |})$$As you can see, you can't ignore the imaginary part of $Z$.(Sorry about the equation spacing but latex won't interpret the "\\\" multiple line instruction today)

 

[fresh_42]

(Sorry about the equation spacing but latex won't interpret the "\\\" multiple line instruction today)

It is accepted within \begin{align*} A&=a \\ B&=b \end{align*}:
\begin{align*}A&=a \\ B&=b\end{align*}

 

[WWGD]

The expression $\sqrt {X+ i\delta}$, will equal a term of the form $a+ib$ , for$a,b$ Real, and then you can write $(1+a) +ib$.

 

[CAF123]

The expression $\sqrt {X+ i\delta}$, will equal a term of the form $a+ib$ , for$a,b$ Real, and then you can write $(1+a) +ib$.

@WWGD thanks, yes this is what I’ve been trying to do. How to write it in such a form if X < 0?

 

[WWGD]

@WWGD thanks, yes this is what I’ve been trying to do. How to write it in such a form if X < 0?

Ok, I I was just thinking of just taking a square root of the expression, which would be of the form a+ib, with neither a nor b equal to 0 and then grouping the parts together. Would that work, or is that too obvious? Edit: I was thinking along the lines of what Fred Wright wrote. Are you looking for an explicit expression for the square root in terms of $X , \delta$?

 

[CAF123]

Ok, I I was just thinking of just taking a square root of the expression, which would be of the form a+ib, with neither a nor b equal to 0 and then grouping the parts together. Would that work, or is that too obvious? Edit: I was thinking along the lines of what Fred Wright wrote. Are you looking for an explicit expression for the square root in terms of $X , \delta$?

Thanks. Yes, I want to find $\sqrt{X+i\delta} = a + ib$ for the cases X <0 and X>0. As $0 < \delta \ll 1$ I can say $b=\pm\delta$ so all I need to decide is 1) the sign of b here and 2) what a is. Perhaps the sign of b depends on whether X is positive or negative but I can’t see how to prove that.

 

[anuttarasammyak]

\sqrt{X+iδ}=(X^2+δ^2)^{1/4} \ e^{iϕ/2}=(X^2+δ^2)^{1/4} (\cos\frac{\phi}{2} + i \ \sin\frac{\phi}{2})
where
ϕ=tan^{−1}\frac{δ}{X}\ for\ X&gt;0
ϕ=π/2\ for\ X=0
ϕ=tan^{−1}\frac{\delta}{X}+π \ for\ X&lt;0
or
ϕ=tan^{−1}\frac{\delta}{X}+\frac{\pi}{2}(1−sgn(X))

As I posted in #2, I wonder why you are so much worrying about the expression with δ though δ→0 does not seem different from $\delta=0$ in its behavior.