I'm just starting in a Complex Analysis course, and I am stuck on a couple questions. The questions are as follows:
If $z = a + bi$ is a point on the curve $|z + 2| + |z - 2| = \sqrt{10},$ find the equation relating $a$ and $b$.
and
Find the roots of $\bar z ^2 + z + 1 = 0.$
Please note that $\bar z$ is referring to the complex conjugate of $z.$
Thanks to anyone who tries to give me a hand! I tried for so long yesterday to solve these, and they're driving me nuts!
Put $z=a+ib$ in $$|z+2|+|z-2| = \sqrt{10}\;,$$ We get $\sqrt{(a+2)^2+b^2}+\sqrt{(a-2)^2+b^2}=\sqrt{10}$
Now Let $$PA=\sqrt{(a+2)^2+b^2}$$ and $$PB = \sqrt{(a-2)^2+b^2}$$
So we get $PA+PB=\sqrt{10}........(1)$
$$\displaystyle (PA+PB)(PA-PB)=\sqrt{10}(PA-PB)\Rightarrow PA-PB = \frac{(PA)^2-(PB)^2}{\sqrt{10}} = \frac{8a}{\sqrt{10}}$$
So we get $$\displaystyle PA-PB = \frac{8a}{\sqrt{10}}.........(2)$$
Now Add $(1)$ and $(2)\;,$ We get
$$\displaystyle 2PA = \sqrt{10}+\frac{8a}{\sqrt{10}} = \frac{10+8a}{\sqrt{10}}\Rightarrow PA=\frac{5+4a}{\sqrt{10}}\Rightarrow (PA)^2= \frac{5+4a}{\sqrt{10}}$$
So we get $$\displaystyle (a+2)^2+b^2 = \frac{5+4a}{\sqrt{10}}\Rightarrow 10\left[(a+2)^2+b^2\right] = (5+4a)^2$$
$(2)$ Given $\bar{z}^2+z=1=0\;,$ Now taking Conjugate, we get $z^2+\bar{z}=1=0$
Now $z^2-\bar{z}^2-(z-\bar{z}) =0\Rightarrow (z-\bar{z})\cdot (z+\bar{z}-1)=0$
So we get $z=\bar{z}$ and $z+\bar{z} = 1$
So $z=\bar{z}$ means $z=a$ Purely Real. and $z+\bar{z}=1$ Means $\displaystyle a=\frac{1}{2}$
$\bullet\; $ If $z=\bar{z}\Rightarrow z=a$ Put into $\bar{z}^2+z=1=0$
we get $a^2+a+1=0$ solving that we get no real value of $a$
So no complex no. $z$ in that case.
$\bullet\; $ If $\displaystyle z+\bar{z} =1 \Rightarrow a=\frac{1}{2}\;,$ Now put $\displaystyle z=\frac{1}{2}+ib$ in $\bar{z}^2+z=1=0$ and solve for $b$