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 $1$?

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    $\begingroup$ The question in the title made no sense, but the body of the post contained a different question which made sense. I've edited the title. $\endgroup$
    – Asaf Karagila
    Commented Feb 26, 2013 at 0:03
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    $\begingroup$ |a| = sqrt(C² + R²) => |1 - i| = sqrt(1² + 1²) = sqrt(2) and |1| = 1. sqrt(2) > 1 => |1 - i| > |1| $\endgroup$
    – njzk2
    Commented Feb 26, 2013 at 11:03

9 Answers 9


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:

$\hspace{5cm}$complex plane

  • 32
    $\begingroup$ I must be very tired if I am thinking about this in terms of geometry. $\endgroup$
    – Asaf Karagila
    Commented Feb 25, 2013 at 23:04
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    $\begingroup$ ...and now I added a drawing. The horror... (If anyone can improve this drawing somehow, please do. :-)) $\endgroup$
    – Asaf Karagila
    Commented Feb 25, 2013 at 23:18
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    $\begingroup$ This is really the best demonstration of this property. +1 $\endgroup$
    – Emily
    Commented Feb 25, 2013 at 23:38
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    $\begingroup$ +1 for being the only reasonable (read: geometric) answer! $\endgroup$
    – Tyler
    Commented Feb 26, 2013 at 3:53
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    $\begingroup$ I concur, given the lack in intuitive understanding of complex numbers the OP showed, this really is the only reasonable answer. $\endgroup$
    – Christian
    Commented Feb 26, 2013 at 10:06

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

  • $\begingroup$ Is this a violation of 7. from here ? 😄 $\endgroup$ Commented May 22, 2019 at 22:51
  • $\begingroup$ @MaximilianJanisch Yes. I evolved :) This is from 2013. I'll fix it. $\endgroup$
    – Git Gud
    Commented May 22, 2019 at 23:00

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


$|1|= 1$

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

Since the modulus is radius in the Argand plane,



Well as far as the title is concerned,




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 $|1-i|\gt1$.


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 $L=\sqrt{2}$


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