A simple Complex Function [Spivak 27-12(a)]

Question

The problem asks to show that for real $x$ we can choose $\log(x+i)$ to be $$\log(x+i)=\log(1+x^2)+i(\frac\pi2-\arctan x).$$ The obvious strategy is to evaluate $x+i$ in the form of its polar coordinates (its absolute value and its argument $\theta$) and then apply the $\log$ function to the resulting equation.

So, the absolute value of a complex number $z=x+iy$ is defined by $\sqrt{x^2+y^2}$; thus in the above case, I thought, I should get the number $\sqrt{x^2+1}$, but instead the answer to the problem claims that it's just $1+x^2$. I don't understand why, what am I missing?

Answer 1

As I commented, Spivak's answer book has a sloppy typographical error. The problem is, in fact, correctly stated, with the square root.

Answer 2

Let $z = x+iy$

Let, $x = r\cos\theta$ and $y = r\sin\theta$

So, $x^2+y^2 = r^2 \implies r = \sqrt{x^2+y^2}$ and $\theta = \arctan\frac{y}{x}$

and $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$

Now,

$\log(x+iy) = \log(r^{i\theta}) = \log(r) + i\theta$

$\log(x+iy) = \log((x^2+y^2)^{1/2}) + i\arctan\frac{y}{x}$

$\log(x+iy) = \frac{1}{2}\log(x^2+y^2) + i\arctan\frac{y}{x}$

For y = 1

$\log(x+i) = \frac{1}{2}\log(x^2+1) + i\arctan\frac{1}{x}$

or

$$\log(x+i) = \frac{1}{2}\log(x^2+1) + i\big(\frac{\pi}{2} - \arctan x\big)$$

As $\tan^{-1}\frac{1}{x} = cot^{-1}{x} = \frac{\pi}{2} - tan^{-1}x$, probably you're missing a $\frac{1}{2}$