complex plane questions

Question

Find where the points of the complex plane are if,

a) $|\pi - \arg z| < \pi/4$

b) $|\Re z| < 1$

c) $\Im \left(\frac{z+1}{z+i}\right) = 0$

d) $z = z_1 + t(\cos x + i\sin x), 0\leq x\leq=\pi/4$ where $z_1 = 1+2i$ and $t=2$

Please kindly help. Thanks.

Answer 1

a) This says the angle for $z$ is between $\frac {3\pi}4$ and $\frac {5\pi}4$.

At least, this is the answer if you are using the generalized arg function, which can give any real answer, especially allowing values between $\pi$ and $2\pi$. If you are using the principal value function Arg, that would limit the angle to between $\frac {3\pi}4$ and $\pi$ and would lose the part of my graph in the third quadrant (negative real and imaginary parts).

b) This says the real part is between $-1$ and $1$.

d) This is a circular arc, centered at $1+2i$ with radius $2$ and angle from $0$ to $\pi/4$.

c) If we let $z=x+yi$ for real $x$ and $y$, this makes $$0=\Im\left(\frac{z+1}{z+i}\right)$$ $$=\Im\left(\frac{x+yi+1}{x+yi+i}\right)$$ $$=\Im\left(\frac{(x+yi+1)(x-yi-i)}{(x+yi+i)(x-yi-i)}\right)$$ $$=\Im\left(\frac{x^2-xyi-xi+xyi+y^2+y+x-yi-i}{x^2+(y+1)^2}\right)$$ $$=\frac{-x-y-1}{x^2+(y+1)^2}$$ $$0=-x-y-1$$ $$y=-x-1$$

The original expression prohibits $z=-i$, so we put an open dot on that point of the graph.