0

complex number question here:

Sketch all complex numbers such that $|z| < |\bar{z} - 1|$, show working and justify answer

Just not sure where to start on this one. Any help would be appreciated (answer even more haha)

SvanN
  • 2,307
Polymer
  • 137

4 Answers4

2

HINT: write $z=x+iy$ and square both sides of your inequality.

Joe
  • 11,745
2

Think geometrically.

$|z|$ expresses the distance between the number in the complex plane and the origin. $|\bar z - 1| = |z - 1|$ expresses the distance between the number and the point 1 in the plane.

For the second part we abused the fact that $|\bar w| = |w|$ to get rid of the conugation operation.

Okay so $|z|<|z - 1|$ denotes the set of points in the plane where the distance to the origin is less than the distance to (1,0).

What does this domain look like?

Squid
  • 1,195
  • From that is the domain basically a unit circle? Just in the complex plane? Or am I barking up the wrong tree here – Polymer Jun 03 '17 at 12:05
  • @Polymer Think of the complex plane. Now think of which points, when we move them one unit to the left, are farther from the origin in the new position than they were in their original position. – Ovi Jun 03 '17 at 12:08
  • Okay so this is where I've got so far, if you could please see if I'm on the right track I'd be very grateful:

    So we let z = x + yi

    Now |z| = sqrt(x^2 + y^2),
    And |z - 1| = sqrt(x^2 + (y - 1)^2), from that we get:
    y < 1/2

    So the domain would be anything under 1/2. Does this have validity or no?

     – Polymer Jun 03 '17 at 12:17
  • Oh no, sorry! Not that at all, since x is the one on the real plane. So the answer would be x < 1/2, yes? – Polymer Jun 03 '17 at 12:24
  • @Polymer. Yes, that's it. – Squid Jun 03 '17 at 15:52
2

An algebraic solution without $z=x+yi$.

The inequality is equivalent to $$ |z|^2<|\bar{z}-1|^2 $$ that can be written $$ z\bar{z}<(\bar{z}-1)\overline{(\bar{z}-1)} $$ and therefore, since $\overline{(\bar{z}-1)}=z-1$, $$ z\bar{z}<\bar{z}{z}-\bar{z}-z+1 $$ that is, cancelling $z\bar{z}$ from both sides and moving terms to the opposite side, which can be done because we only act on real numbers, $$ z+\bar{z}<1 $$ Can you write $z+\bar{z}$ in a different way?

egreg
  • 238,574
1

$\overline{z}$ is a complex conjugate of z ($\overline{a+bi} = a – bi$). $|z| < |\overline{z} – 1| <=> |Re(z) + Im(z)i| < |Re(z) – Im(z)i – 1| <=> {|Re(z) + Im(z)i|}^2 <{|Re(z) – 1 – Im(z)i|}^2 <=> {Re(z)}^2 + {Im(z)}^2 < {Re(z) - 1}^2 + {Im(z)}^2 <=> {Re(z)}^2 < {(Re(z) - 1)}^2 <=> 0 < 1 – 2Re(z) <=> Re(z) < \frac{1}{2}$

Chain Markov
  • 15,564
  • 6
  • 36
  • 116