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 $1$?
 A: One of the properties of complex numbers is that we cannot compare them, but we can compare their modulus. 
you can compare $|1-i|$ with $|1|$, equivalent with $\sqrt{1^{2}+1^{2}}\geq 1.$
A: By definition, the absolute value of a given complex number $z$ with $z=x+iy$ (for some $x,y\in \Bbb R$) is $\sqrt{x^2+y^2}$ and it is denoted by $|z|$.
If $z=1-i$, according to the definition we get $|1-i|=\sqrt{1^2+(-1)^2}=\sqrt {2}>1=|1|$.
A: $|1|= 1$
$|1-i|= \sqrt{1^2+(-1)^2}=\sqrt{2}$
Since the modulus is radius in the Argand plane,
$|1-i|>|1|$
A: Hint: By definition, if $a+bi$ is a complex number ($a$ and $b$ being real numbers, as usual),
$$|a+bi|=\sqrt{a^2+b^2}.$$

Also, it is really incorrect to say that "$1-i$ is larger than $1$"; the complex numbers have no ordering. You should just instead say that $|1-i|>1$.
A: Well as far as the title is concerned,
$|1-i|=\sqrt2$>$|1|$=1
$|1-i|$>$|1|$
A: This is a small variant of Asaf Karagila's geometric proof:  If $a$ and $b$ are two points in the complex plane, the distance between them is $|a-b|$, so if you draw the (right isosceles) triangle with $a=1$, $b=i$ and $c=0$, you see that $|1-i|\gt1$.
A: In complex number, you have to compute the module of a number to say if it is "larger" than something. In general the module is $$|a+bi|=\sqrt{a^2+b^2}$$
In your case 
$$|1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}$$
Then 
$$\sqrt{2}\geq|1|$$
And this imply that $1-i$ lies outside of the unit circle
A: 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 $\sqrt2>1$.
Here is a drawing:
$\hspace{5cm}$
A: 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 $L=\sqrt{2}$
