Finding which complex numbers satisfy $\left|z + 1\right| + \left|z - 1\right| = 4$ This problem is from the book Complex Analysis with Applications by Asmar. Currently I am stuck at, well, everything, as I don't know know how to manipulate the expression $\left|z + 1\right| + \left|z - 1\right| = 4$ into anything meaningful. Squaring and simplifying both sides yields $z\bar{z} + \left|z + 1\right|\left|z - 1\right| + 1 = 8$, and I don't see how that is useful.
The problem is asking us to find those complex numbers whose sum of distances from both $1$ and $-1$ is equal to $4$. I suppose the end result is a circle of some kind, as points $(2, 0)$ and $(-2, 0)$ satisfy the equality.
 A: If you write $z$ as $a+bi$ with $a,b\in\Bbb R$, then\begin{align}|z+1|+|z-1|=4&\iff\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}=4\\&\iff2a^2+2b^2+2+2\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=16\\&\iff\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\\&\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\\&\iff3a^2+4b^2=12.\end{align}So, you get the ellipse $3a^2+4b^2=12$.
However, note that the equivalence\begin{multline}\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\iff\\\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\end{multline}is not obvious, but is not hard to prove that $|z-1|+|z+1|=4\implies|z|^2\leqslant7$.
A: Let $z = x + iy$, then $|z+1| = \sqrt{(x+1)^2 + y^2}$ and $|z-1| = \sqrt{(x-1)^2 + y^2}$.
$|z+1| = 4 - |z-1|$
$(x+1)^2 + y^2 = 16 + (x-1)^2 + y^2 - 8\sqrt{(x-1)^2 + y^2}$
$x - 4 = -2\sqrt{(x-1)^2 + y^2}$
$x^2 + 16 - 8x = 4(x-1)^2 + 4y^2$
$12 = 4y^2 + 3x^2$
$\frac{x^2}{4} + \frac{y^2}{3} = 1$ which is formula of ellipse.
A: The given equation tells us that the foci of ellipse are $(1,0)$ and $(-1,0)$. Symmetry tells us that the center of the ellipse is $(0,0)$. Also, $4$ on the RHS of the given equation is the length of the major axis. So, the length of semi major axis, $a=2$. Also, $ae=1\implies e=\frac12$ is the eccentricity of the ellipse.
Now, the length of the semi minor axis, $b$ can be calculated using $b^2=a^2(1-e^2)=4(1-\frac14)=3$
Putting  these values in the equation of ellipse i.e. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get $\frac{x^2}{4}+\frac{y^2}{3}=1$
A: Choose z = r * e^(i * phi); insert this in the given equation, and solve for r = r(phi).
This will give you r (phi) = (2 Sqrt[3])/Sqrt[3 Cos[phi]^2 + 4 Sin[phi]^2], the equation of an ellipse with a = 2 (on the real axis) and b = sqrt[3] (on the imaginary axis), and the origin as the center.
Any z on this ellipse will satisfy the given equation.
