Chapter 4 - Complex Numbers and Quadratic Equations

Let $z = x + iy$. Then $|z + 1| = |x + 1 + iy| = \sqrt{(x+1)^2 + y^2}$.
Given equation: $\sqrt{(x+1)^2 + y^2} = x + iy + 2 + 2i = (x+2) + i(y+2)$.
Comparing imaginary parts: $0 = y + 2 \implies y = -2$.
Comparing real parts: $\sqrt{(x+1)^2 + (-2)^2} = x + 2$.
$\sqrt{x^2 + 2x + 1 + 4} = x + 2$.
Squaring both sides: $x^2 + 2x + 5 = (x+2)^2 = x^2 + 4x + 4$.
$2x + 5 = 4x + 4 \implies 2x = 1 \implies x = \frac{1}{2}$.
Wait, let's recheck. $x^2+2x+5 = x^2+4x+4 \implies 2x = 1 \implies x=1/2$.
Let's check snippet solution again. Snippet said $x=3/2$. Let's re-calculate.
Snippet equation: $|z+1| = z + 2(1+i)$.
Maybe snippet meant $|z| = z + 1 + 2i$ (Q11).
Let's check Q12 text again. "If $|z+1| = z + 2(1+i)$"
My calculation: $x=1/2, y=-2$. Let's check validity.
LHS: $|1/2 - 2i + 1| = |3/2 - 2i| = \sqrt{9/4 + 4} = \sqrt{25/4} = 5/2$.
RHS: $1/2 - 2i + 2 + 2i = 5/2$. Matches.
So $z = \frac{1}{2} - 2i$.

Given $\left| \frac{z_1 - 2z_2}{2 - z_1 \bar{z}_2} \right| = 1$. Squaring both sides:
$\left| z_1 - 2z_2 \right|^2 = \left| 2 - z_1 \bar{z}_2 \right|^2$
$(z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2) = (2 - z_1 \bar{z}_2)(2 - \bar{z}_1 z_2)$
$|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + 4|z_2|^2 = 4 - 2\bar{z}_1z_2 - 2z_1\bar{z}_2 + |z_1|^2|z_2|^2$
$|z_1|^2 + 4|z_2|^2 = 4 + |z_1|^2|z_2|^2$
$|z_1|^2 - |z_1|^2|z_2|^2 + 4|z_2|^2 - 4 = 0$
$|z_1|^2(1 - |z_2|^2) - 4(1 - |z_2|^2) = 0$
$(|z_1|^2 - 4)(1 - |z_2|^2) = 0$
Since $|z_2| \neq 1$, we have $|z_1|^2 - 4 = 0 \implies |z_1| = 2$.

LHS$^2 = \left| \frac{z_1 - z_2}{1 - \bar{z}_1 z_2} \right|^2 = \frac{(z_1 - z_2)(\bar{z}_1 - \bar{z}_2)}{(1 - \bar{z}_1 z_2)(1 - z_1 \bar{z}_2)}$
$= \frac{|z_1|^2 - z_1\bar{z}_2 - \bar{z}_1z_2 + |z_2|^2}{1 - z_1\bar{z}_2 - \bar{z}_1z_2 + |z_1|^2|z_2|^2}$
Since $|z_1| = 1$, $|z_1|^2 = 1$.
Numerator $= 1 - z_1\bar{z}_2 - \bar{z}_1z_2 + |z_2|^2$.
Denominator $= 1 - z_1\bar{z}_2 - \bar{z}_1z_2 + 1 \cdot |z_2|^2$.
Numerator equals Denominator, so the ratio is 1.

LHS $= (1 - \bar{z}_1 z_2)(1 - z_1 \bar{z}_2) - (z_1 - z_2)(\bar{z}_1 - \bar{z}_2)$
$= (1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + |z_1|^2|z_2|^2) - (|z_1|^2 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + |z_2|^2)$
$= 1 + |z_1|^2|z_2|^2 - |z_1|^2 - |z_2|^2$
$= 1(1 - |z_2|^2) - |z_1|^2(1 - |z_2|^2)$
$= (1 - |z_1|^2)(1 - |z_2|^2)$ = RHS.

Let $z_1 = r e^{i\theta_1}$ and $z_2 = r e^{i\theta_2}$ (since moduli are equal).
Given $\theta_1 + \theta_2 = \pi \implies \theta_1 = \pi - \theta_2$.
$z_1 = r e^{i(\pi - \theta_2)} = r (\cos(\pi - \theta_2) + i \sin(\pi - \theta_2))$
$= r (-\cos \theta_2 + i \sin \theta_2)$.
Now, $\bar{z}_2 = r (\cos \theta_2 - i \sin \theta_2)$.
$-\bar{z}_2 = r (-\cos \theta_2 + i \sin \theta_2)$.
Thus, $z_1 = -\bar{z}_2$.

Since $|z_k| = 1$, we have $z_k \bar{z}_k = 1 \implies \bar{z}_k = \frac{1}{z_k}$.
RHS $= \left| \bar{z}_1 + \bar{z}_2 + \dots + \bar{z}_n \right|$
$= \left| \overline{z_1 + z_2 + \dots + z_n} \right|$
We know that $|\bar{z}| = |z|$.
So, RHS $= |z_1 + z_2 + \dots + z_n|$ = LHS.

Given $z_2 = \bar{z}_1$ and $z_4 = \bar{z}_3$.
Expression $= \arg(z_1) - \arg(z_4) + \arg(z_2) - \arg(z_3)$
$= \arg(z_1) - \arg(\bar{z}_3) + \arg(\bar{z}_1) - \arg(z_3)$
$= \arg(z_1) - (-\arg(z_3)) + (-\arg(z_1)) - \arg(z_3)$
$= \arg(z_1) + \arg(z_3) - \arg(z_1) - \arg(z_3) = 0$.

Using the result from Q17, since $|z_k| = 1$, we have $|z_1 + z_2 + z_3| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right|$.
Given the RHS is 1, the value of $|z_1 + z_2 + z_3|$ is also 1.

Given $|9z_1z_2 + 4z_1z_3 + z_2z_3| = 12$.
Factor out $z_1z_2z_3$: $\left| z_1z_2z_3 \left( \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right) \right| = 12$.
$|z_1||z_2||z_3| \left| \frac{9\bar{z}_3}{|z_3|^2} + \frac{4\bar{z}_2}{|z_2|^2} + \frac{1\bar{z}_1}{|z_1|^2} \right| = 12$.
Substitute moduli: $1 \cdot 2 \cdot 3 \left| \frac{9\bar{z}_3}{9} + \frac{4\bar{z}_2}{4} + \frac{\bar{z}_1}{1} \right| = 12$.
$6 |\bar{z}_3 + \bar{z}_2 + \bar{z}_1| = 12$.
$6 |\overline{z_1 + z_2 + z_3}| = 12$.
$6 |z_1 + z_2 + z_3| = 12 \implies |z_1 + z_2 + z_3| = 2$.

Let $z = x + iy$. $|z| = 2 \implies x^2 + y^2 = 4$.
$\arg(z+1) = \frac{\pi}{4} \implies \tan^{-1}\left(\frac{y}{x+1}\right) = 1$.
$\frac{y}{x+1} = 1 \implies y = x + 1$ (with $x+1 > 0, y > 0$).
Substitute $y$ in circle equation: $x^2 + (x+1)^2 = 4$.
$x^2 + x^2 + 2x + 1 = 4 \implies 2x^2 + 2x - 3 = 0$.
$x = \frac{-2 \pm \sqrt{4 - 4(2)(-3)}}{4} = \frac{-2 \pm \sqrt{28}}{4} = \frac{-1 \pm \sqrt{7}}{2}$.
Since $y = x+1 > 0$, we need $x > -1$. $\frac{-1 - \sqrt{7}}{2} < -1$, so reject.
$x = \frac{-1 + \sqrt{7}}{2}$. Then $y = \frac{1 + \sqrt{7}}{2}$.
Wait, let's check if there's a simpler integer solution often found in Exemplar.
Maybe the question is different. Let's check snippet [20] again. "If $z_1 \neq -z_2$ and $|z_1+z_2| = |1/z_1 + 1/z_2|$..."
Let's stick to the list I derived. I will use the "Solve $z + \sqrt{2}|z+1| + i = 0$" as Q22 and leave Q21 as this one or similar.

Let $z = x + iy$.
$x + iy + \sqrt{2}\sqrt{(x+1)^2 + y^2} + i = 0$.
Equating imaginary parts: $y + 1 = 0 \implies y = -1$.
Equating real parts: $x + \sqrt{2}\sqrt{(x+1)^2 + (-1)^2} = 0$.
$\sqrt{2}\sqrt{x^2 + 2x + 2} = -x$.
Squaring (condition $-x \ge 0 \implies x \le 0$):
$2(x^2 + 2x + 2) = x^2$.
$2x^2 + 4x + 4 = x^2 \implies x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2$.
Since $x = -2 \le 0$, it is a valid solution.
So $z = -2 - i$.

Numerator: $1 - i = \sqrt{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = \sqrt{2} \left( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right) = \sqrt{2} e^{-i\pi/4}$.
Denominator: $e^{i\pi/3}$.
$z = \frac{\sqrt{2} e^{-i\pi/4}}{e^{i\pi/3}} = \sqrt{2} e^{-i(\pi/4 + \pi/3)} = \sqrt{2} e^{-i(7\pi/12)}$.
Polar form: $\sqrt{2} \left( \cos \frac{7\pi}{12} - i \sin \frac{7\pi}{12} \right)$.

Let $z = r_1 e^{i\theta_1}$ and $w = r_2 e^{i\theta_2}$.
$|zw| = r_1 r_2 = 1$.
$\arg(z) - \arg(w) = \theta_1 - \theta_2 = \frac{\pi}{2} \implies \theta_2 = \theta_1 - \frac{\pi}{2}$.
$\bar{z}w = (r_1 e^{-i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_2 - \theta_1)}$.
$= 1 \cdot e^{i(-\pi/2)} = \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) = 0 - i = -i$.

Let $z = \frac{2-i}{(1-2i)^2}$.
$(1-2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i$.
$z = \frac{2-i}{-3-4i} = \frac{2-i}{-(3+4i)} \cdot \frac{3-4i}{3-4i}$
$= \frac{6 - 8i - 3i + 4i^2}{-(9+16)} = \frac{6 - 11i - 4}{-25} = \frac{2 - 11i}{-25} = -\frac{2}{25} + i\frac{11}{25}$.
Conjugate $\bar{z} = -\frac{2}{25} - i\frac{11}{25}$.

No. For example, let $z_1 = 1$ and $z_2 = i$.
$|z_1| = 1$ and $|z_2| = 1$, so $|z_1| = |z_2|$.
But $z_1 \neq z_2$.

$x^2 + y^2 = |x+iy|^2 = \left| \frac{(a+i)^2}{2a-i} \right|^2$
$= \frac{|a+i|^4}{|2a-i|^2} = \frac{(\sqrt{a^2+1})^4}{(\sqrt{4a^2+1})^2} = \frac{(a^2+1)^2}{4a^2+1}$.

$z = |z|(\cos \theta + i \sin \theta) = 4(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6})$
$= 4(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -2\sqrt{3} + 2i$.

Numerator: $(1+i)(2+i) = 2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i$.
Expression $= \frac{1+3i}{3+i} \cdot \frac{3-i}{3-i} = \frac{3 - i + 9i - 3i^2}{9+1} = \frac{3 + 8i + 3}{10} = \frac{6+8i}{10} = \frac{3}{5} + \frac{4}{5}i$.

Let $z = 1 + i\sqrt{3}$. $|z| = 2, \arg(z) = \frac{\pi}{3}$.
$z^2 = (2 e^{i\pi/3})^2 = 4 e^{i2\pi/3}$.
Principal argument is $\frac{2\pi}{3}$.

$|\frac{z-5i}{z+5i}| = 1 \implies |z-5i| = |z+5i|$.
This represents the locus of points equidistant from $5i$ and $-5i$.
The perpendicular bisector of the segment joining $(0, 5)$ and $(0, -5)$ is the line $y = 0$, which is the real axis.

(a) Let $z = \sqrt{3} + i$. $r = 2, \tan \alpha = 1/\sqrt{3} \implies \alpha = \pi/6$. Polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$. Matches (v).
(b) $z = -1 + i\sqrt{3}$. 2nd quadrant. $\alpha = \pi/3$. Arg $= \pi - \pi/3 = 2\pi/3$. Matches (iii).
(c) $|z - (-2)| = |z - 2|$. Perpendicular bisector of segment joining $(-2, 0)$ and $(2, 0)$. Matches (i).
(d) $|z - (-2i)| = |z - 2i|$. Perpendicular bisector of segment joining $(0, -2)$ and $(0, 2)$. Matches (iv).
(e) $|z - (-4i)| \ge 3$. Distance from $(0, -4)$ is $\ge 3$. On or outside circle. Matches (ii).
(f) $|z - (-4)| \le 3$. Distance from $(-4, 0)$ is $\le 3$. On or inside circle. Matches (vi).
(g) $z = \frac{1+2i}{1-i} = \frac{(1+2i)(1+i)}{2} = \frac{1+3i-2}{2} = -\frac{1}{2} + \frac{3}{2}i$. Conjugate $\bar{z} = -\frac{1}{2} - \frac{3}{2}i$. 3rd quadrant. Matches (viii).
(h) $1/(1-i) = (1+i)/2 = 1/2 + i/2$. 1st quadrant. Matches (vii).