# Sketch $z\in \mathbb{C}:0 < arg(z-(1+i)) < \frac\pi3$

Sketch the following $$z\in \mathbb{C}:0 < arg(z-(1+i)) < \frac\pi3$$ I have considered this geometrically and ended up thinking that the complex numbers $z$ must satisfy $$0 < \frac{Im(z-(1+i))}{Re(z-(1+i))} < \sqrt{3}$$ But this doesn't seem to help. I am thinking im approaching the problem in the wrong way. What is the best way to attempt to sketch this region of $\mathbb{C}$ this without using a computer? Thanks

here is a way to plot the region. think of the point $1+i$ as the origin. draw two half rays originating at $1+i$ and one going parallel to the $x$-axis and in the positive direction and the other one at an angle $60^\circ.$ the wedge between the rays is the region you are after.
• I think I understand what you mean. By 'rays' do you mean just a straight line? and the two of them make a wedge shape with $1+i$ at the tip? What was your logic behind that – Trawkley Jan 31 '15 at 22:02
• @Trawkley, by a i mean it goes in only one direction. not line through the point. the point $i+i$ is at the end of both lines(half rays) not in the middle if you drew a line infinite in both directions. – abel Jan 31 '15 at 22:06