1
$\begingroup$

There are infinitely many complex numbers $z$ such that $|z|= 1$. Can anybody just explain this to me so I can get the picture.

$\endgroup$
12
$\begingroup$

enter image description here

Here any $z$ on this circle satisfy $|z|=1$

Here $z=a+ib$ ie. $z=(a,b) $ and can be represented as a point or vector on complex plane above. $|z|^2=a^2+b^2 =1$. and this itself is a locus of a circle.

$\endgroup$
  • 1
    $\begingroup$ would you mind if I draw your graphic in TikZ ? yours look so much like paint. $\endgroup$ – Dominic Michaelis Apr 8 '13 at 13:28
  • $\begingroup$ @DominicMichaelis I would not mind but keep it simple as this one $\endgroup$ – ABC Apr 8 '13 at 16:13
  • $\begingroup$ ok made it, if i shall change anything tell me $\endgroup$ – Dominic Michaelis Apr 8 '13 at 16:59
  • $\begingroup$ fixed it even changed norm with absolute value $\endgroup$ – Dominic Michaelis Apr 8 '13 at 17:08
  • $\begingroup$ @DominicMichaelis Better than never. Thanx. $\endgroup$ – ABC Apr 8 '13 at 17:09
6
$\begingroup$

Let $$z = a + ib$$

Now $$|z| = |a + ib|$$ $$ |z|^2 = a^2+b^2$$ $$ |z| = \sqrt{a^2 + b^2}$$

Hence given equation becomes, $$a^2 + b^2 = 1$$

which is the equation of a circle. Hence there are infinitely many points on this unit circle which satisfy the given equation.

$\endgroup$
  • $\begingroup$ Thanks that makes things clear. $\endgroup$ – Lynnie Apr 8 '13 at 12:27
  • 2
    $\begingroup$ Just a small point, $|z|^2 = a^2 + b^2$, not $|z|$. $\endgroup$ – muzzlator Apr 8 '13 at 12:46
0
$\begingroup$

There are not only infinitely many complex numbers on the unit circle, there are infinitely many Gaussian numbers (both real and imaginary parts being rational) there. For, take a Gaussian integer $z=m+ni$, both $m$ and $n$ being ordinary integers, and calculate $z/\bar z$, something that you should have done many times in high school. You see that the resulting complex number $x+iy$ has both $x$ and $y$ in $\mathbb Q$, and of course its absolute value is $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.