Greatest value of $|z|$ such that $|z-2i|\le2$ and $ 0\le \arg(z+2)\le 45^\circ$

Question

I've to sketch the complex numbers $z$ satisfying both the inequalities

$$|(z-2i)|\le2,$$ $$ 0\le \arg(z+2)\le 45^\circ.$$

I was able to sketch and shade the region that satisfies both inequalities; here is my Argand diagram:

However, I've a problem in getting the greatest value of $|z|$, i.e. the maximum length of $z$ from $(0,0).$ It should come to around 3.70.

Answer 1

if you parametrize the circle of constraint it is $(2\cos t,2\sin t +2)$ with $t=0$ corresponding to the point $(2,2)$. so:

$$ \frac18 |z|^2 = \frac12 (\cos^2 t +\sin^2 t + 2 \sin t +1 )= 1+ \sin t $$

Answer 2

Inspection of the figure indicates that the point $z_0:=\sqrt{2}+i(2+\sqrt{2})$ is the feasible point with largest absolute value: The closed disk $D_0$ with center $0$ and radius $|z_0|$ obviously contains the complete feasible region $F$, and $z_0\in F$. It follows that $$\max_{z\in F}|z|=|z_0|=\sqrt{8+4\sqrt{2}}\doteq3.6955\ .$$

Answer 3

Let $0+2i$ be C (the centre of the circle), let the point of furthest intersection (greatest $|z|$) of the 2 loci be A (You have indicated it on your diagram with a cross), construct the point B such that $\triangle CAB$ has a right angle at B.

From here we can note that $\angle CAB=\frac{\pi}{4}$(given - construction of CB is such that it is parallel to the x-axis, use corresponding angles), $AC=2$ (radii). Using trigonometry we can find that $BC=2\cos(\frac{\pi}{4})=\sqrt{2}$, similarly $AB=2\sin(\frac{\pi}{4})=\sqrt{2}$.

Applying our knowledge to the diagram (the Argand diagram), we see that the real part of the desired complex number is given by $BC$, $\Re(z)=\sqrt{2}$. The imaginary part is the sum of $AB$ and the height (distance) from the number $\sqrt{2}$ to $\sqrt{2}+2i$ (distance is $2$), thus $\Im(z)=BC+2=2+\sqrt{2}$.

$\therefore$ $z=\sqrt{2}+i(2+\sqrt{2})$ and $|z|=2\sqrt{2+\sqrt{2}}$

Answer 4

If you havent solved it yet, heres my solution. Find the equation of the circle and the straight line through the circle. eqn of circle is:$x^2 + (y-2)^2 = 2^2$ eqn of line is $y= x+2$ There you can see the greatest value of $|z|$ is the intersection point of the circle and the line you drew. Substitute the line equation into the circle equation to find the coordinates of the point of intersection The greatest value of $|z|$ would be the distance of the origin to the point you just found, calculated using the distance formula

Answer 5

First the region $\Delta$ we are considering is the intersection of the three subsets of $\Bbb{C}$: $D$, closed region inside the circle having center $2i$ and radius $2$; $H_1$ closed half-plane of equation $y⩽x$; and $H_2$ closed half-plane of equation $y⩾0$. (Since $D\cap H_1\subset H_2$ we have $\Delta=D\cap H_1$ only.) As such, $\Delta$ is closed & bounded (compact…), and $|z|$ attains its maximum on it; also this maximum can't be attained at an interior point. We must then search the maximum of $|z|$ on the boundary, that is the reunion of a semicircle $\mathscr{C}_-$ and a segment $\mathscr{S}$.

On the semicircle. Let $\theta\in[0,\pi]$ be an argument for $z$; the equation $|z-2i|=2$ is equivalent to $z=(4\sin\theta) e^{i\theta}$. On the other hand, the condition $\arg(z+2)\in\bigl[0,\frac{\pi}4\bigr]$ may be written $0⩽\frac{y}{x+2}⩽1$, equivalent to a $2^{\rm nd}$ degree inequality in $\tan\theta$ which is: $\tan^2\theta-2\tan\theta-1⩽0$. This is equivalent to $\tan\theta\in [1-\sqrt2\mathbin{,}1+\sqrt2]$, or $\theta\in\bigl[0\mathbin{,}\frac{3\pi}8\bigr]\cup \bigl[\frac{7\pi}8\mathbin{,}\pi\bigr]$. Obviously $|z|=4\sin\theta$ attains its maximum over this last set when $\theta=\frac{3\pi}8$, which provides $z=4\sin\bigl(\frac{3\pi}8\big)\exp\bigl( i\frac{3\pi}8 \bigr)=\sqrt2 + i(2+\sqrt2)$ and $|z|=4\sin\bigl(\frac{3\pi}8\bigr)=2\sqrt{2+\sqrt2}\approx3.696$.

On the segment $\mathscr{S}$. The coordinates of the extremities satisfy both $y=x+2$ and $x^2+(y-2)^2=4$, which immediately leads to $x=±\sqrt2$. Thus $\mathscr{S}=\bigl\{x+(x+2)i\quad \big|\quad -\sqrt2⩽x⩽\sqrt2\bigr\}$.

Now $|z|^2=x^2+(x+2)^2=2(x+1)^2+2=g(x)$. This function strictly decreases before $-1$ and strictly increases after $-1$. It is now easy to see that $g$ attains its maximum on $\bigl[-\sqrt2\mathbin{,}\sqrt2\bigr]$ at $\sqrt2$, that is for $z=\sqrt2 + i(2+\sqrt2)$, the value already found on the semicircle.