Nonsensical result that $\lim_{a\to\infty}\mathfrak{Re}(\sqrt{ia + b}) = \infty$ for real $a$, $b > 0$

Question

Let $b > 0$ be a fixed real constant and take $z_a = ia + b$. Then $\mathfrak{Re}(z_a) = b$ for all $a\in\mathbb{R}$. I have managed to confuse myself by reasoning that $\lim_{a\to\infty}\mathfrak{Re}(\sqrt{z_a}) = \infty$ in the following way:

Write $w_a = \sqrt{z_a} = \sqrt{ia + b}$. What is the asymptotics of $\mathfrak{Re}(w_a)$ as $a\to \pm\infty$? $z_a$ form the vertical line intersecting the real line at $b$ and $z_a$'s principal argument clearly satisfies

$$\mathrm{Arg}(z_a)\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$$

for all $a$. This would then lead me to believe that $\mathfrak{Re}(w_a)$ is finite for all $a$, but

$$w_a = \sqrt{|w_a|}e^{i\mathrm{Arg}(w_a)} = \sqrt{|ia + b|}e^{i\mathrm{Arg}(z_a)/2}$$

and $|ai + b| = \sqrt{a^2 + b^2}$. Thus

$$w_a = \left(a^2 + b^2\right)^{1/4}e^{i\mathrm{Arg}(w_a)/2}$$

so that

$$\mathfrak{Re}(w_a) = \left(a^2 + b^2\right)^{1/4}\cos\left(\mathrm{Arg}(z_a)/2\right)$$

Since $\mathrm{Arg}(z_a)/2\in \left(\frac{-\pi}{4},\frac{\pi}{4}\right)$ for all $a$, then $\cos\left(\mathrm{Arg}(z_a)/2\right) > 2^{-1/2}$ and so

$$\lim_{a\to\infty}\mathfrak{Re}(w_a) = \infty$$

which seems nonsensical. What is going on?

Answer 1

Here is the graph of $\operatorname{Re}\big(\sqrt{ia+1}\big)$. Yes, it does go to $\infty$ as $a \to \infty$.

Why do you say it is nonsensical? In fact $$ \lim_{a\to\infty} \frac{\operatorname{Re}\big(\sqrt{ia+1}\big)}{\sqrt{a}} = \frac{1}{\sqrt2} $$

Answer 2

We have that

$$z=b+ai=|z|e^{i\theta} $$

with $\theta=\arctan \frac a b\in \left(-\frac \pi 2, \frac \pi 2\right) \to \operatorname{sign}(a)\cdot\frac \pi 2$ and then

$$z^\frac12 = |z|^\frac12e^{i\left(\frac{\theta}2+k\pi\right)} $$

with $k=0,1$ therefore

• for $k=0$

$$\left(\frac{z}{|z|}\right)^\frac12=e^{i\frac{\pi}4\operatorname{sign}(a)} \to \frac{\sqrt2}2(1+i\operatorname{sign}(a))$$

• for $k=1$

$$\left(\frac{z}{|z|}\right)^\frac12=e^{i\frac{5\pi}4\operatorname{sign}(a)} \to \frac{\sqrt2}2(-1-i\operatorname{sign}(a))$$