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.

$\endgroup$
0

3 Answers 3

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.

$\endgroup$
5
  • 1
    $\begingroup$ would you mind if I draw your graphic in TikZ ? yours look so much like paint. $\endgroup$ Apr 8, 2013 at 13:28
  • $\begingroup$ @DominicMichaelis I would not mind but keep it simple as this one $\endgroup$
    – ABC
    Apr 8, 2013 at 16:13
  • $\begingroup$ ok made it, if i shall change anything tell me $\endgroup$ Apr 8, 2013 at 16:59
  • $\begingroup$ fixed it even changed norm with absolute value $\endgroup$ Apr 8, 2013 at 17:08
  • $\begingroup$ @DominicMichaelis Better than never. Thanx. $\endgroup$
    – ABC
    Apr 8, 2013 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.

$\endgroup$
2
  • $\begingroup$ Thanks that makes things clear. $\endgroup$
    – Lynnie
    Apr 8, 2013 at 12:27
  • 3
    $\begingroup$ Just a small point, $|z|^2 = a^2 + b^2$, not $|z|$. $\endgroup$
    – muzzlator
    Apr 8, 2013 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 $1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .