Is $|1-i|$ larger than $|1|$?

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 $\sqrt2>1$.

Here is a drawing:

$\hspace{5cm}$

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 $|1-i|=\sqrt{1^2+(-1)^2}=\sqrt {2}>1=|1|$.

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 $|1-i|>1$.

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

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 $\sqrt{1^{2}+1^{2}}\geq 1.$