# Complex numbers $z$ such that $|z|= 1$

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

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.

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

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.

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

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 $1$.