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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?

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  • $\begingroup$ You know, it is more than a bit silly to omit the book title when giving a book reference. $\endgroup$ – rschwieb May 22 at 16:34
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As I commented, Spivak's answer book has a sloppy typographical error. The problem is, in fact, correctly stated, with the square root.

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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 $\frac{1}{2}$

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  • $\begingroup$ I get that, but that doesn't exactly answer my question: why does the author claim that $|x+i|=1+x^2$? $\endgroup$ – Simone May 22 at 16:35
  • $\begingroup$ Magnitude of a complex number is defined like that. For reference, en.wikipedia.org/wiki/Complex_number $\endgroup$ – Ak19 May 22 at 16:36
  • $\begingroup$ It's still not clear to me why the absolute value of $x+i$ should be $1+x^2$ instead of $\sqrt{1+x^2}$ $\endgroup$ – Simone May 22 at 16:44
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    $\begingroup$ Quite simply, there is a typo in the answer book. Of course, $|x+i| = \sqrt{1+x^2}$. $\endgroup$ – Ted Shifrin May 22 at 17:06
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    $\begingroup$ Yes, that's a typo, thanks:) $\endgroup$ – Ak19 May 23 at 5:13

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