Is $|1-i|$ larger than $|1|$? I am confused about complex numbers. Does $1-i$ lie outside the unit circle? How do I show that the absolute value of $1-i$ is larger than that of $1$?
Let me do something very different for me. Let me give a geometric proof:

Note that $1-i$ is the bottom-right corner of the square whose center is $0$ and whose edges have length $2$. The other corners are $1+i; -1+i; -1-i$.

The diagonal running from $-1+i$ to $1-i$ is a straight line passing through $0$. Its length, by the Pythagoras theorem, is $2\sqrt2=\sqrt8$. Therefore the distance between $0$ and each of the corners is exactly half, i.e. $\sqrt2$.

And it is trivial to see that $\sqrt2&gt;1$.

Here is a drawing:

$\hspace{5cm}$
Here is another different approach:

The unique angle (in $[0,2\pi)$) between the vector $(1,i)$ in the complex plane and the x-axis is $\phi=\frac{7\pi}{4}$

Euler's identity yields now: 

$$1-i=L(e^{\frac{7\pi}{4}i})=L\left(\cos\left(\frac{7\pi}{4}\right)+i\sin\left(\frac{7\pi}{4}\right)\right)=L\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)=L\left(\cos\left(\frac{\pi}{4}\right)-i\sin\left(\frac{\pi}{4}\right)\right)=L\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right) $$

Hence $L=\sqrt{2}$