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

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?

• You know, it is more than a bit silly to omit the book title when giving a book reference. – rschwieb May 22 at 16:34

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

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 $$\frac{1}{2}$$

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