Chapter 4 - Complex Numbers and Quadratic Equations

Given $z = \bar{w}$. $\sin x + i \cos 2x = \overline{\cos x - i \sin 2x} = \cos x + i \sin 2x$ Comparing real and imaginary parts: $\sin x = \cos x \implies \tan x = 1 \implies x = n\pi + \frac{\pi}{4}$ $\cos 2x = \sin 2x \implies \tan 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8}$ There is no common value of $x$ satisfying both conditions simultaneously.

Let $z = \frac{1-i\sin\alpha}{1+2i\sin\alpha}$. Multiply numerator and denominator by $1-2i\sin\alpha$: $z = \frac{(1-i\sin\alpha)(1-2i\sin\alpha)}{1+4\sin^2\alpha} = \frac{1-2\sin^2\alpha - 3i\sin\alpha}{1+4\sin^2\alpha}$ For $z$ to be purely real, the imaginary part must be zero: $-3\sin\alpha = 0 \implies \sin\alpha = 0 \implies \alpha = n\pi$.

Given $z$ is in 3rd quadrant, so $x < 0, y < 0$. $\frac{\bar{z}}{z} = \frac{x-iy}{x+iy} = \frac{(x-iy)^2}{x^2+y^2} = \frac{x^2-y^2-2ixy}{x^2+y^2}$ For $\frac{\bar{z}}{z}$ to be in 3rd quadrant: Real part $< 0 \implies x^2 - y^2 < 0 \implies x^2 < y^2 \implies |x| < |y|$. Since $x,y$ are negative, this means $y < x$. Imaginary part $< 0 \implies -2xy < 0 \implies xy > 0$, which is true since $x,y$ are both negative. Thus, $y < x < 0$.

$(z+3)(\bar{z}+3) = (z+3)(\overline{z+3}) = |z+3|^2$.

$\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2+1^2} = \frac{1-1+2i}{2} = i$. So, $i^x = 1$. This implies $x$ must be a multiple of 4, i.e., $x = 4n$.

Since $x^2+y^2=1$, let $z = x+iy$, then $|z|=1$ and $\bar{z} = 1/z$. The expression is $\frac{1+z}{1+\bar{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{1+z}{\frac{z+1}{z}} = z = x+iy$.

Clockwise rotation by $\pi/2$ corresponds to multiplication by $e^{-i\pi/2} = -i$. New position $z' = (2-i)(-i) = -2i + i^2 = -1-2i$.

A complex number is non-real if its imaginary part is non-zero. Hence $y \neq 0$.

If two complex numbers are equal, their moduli are equal. $|a+ib| = |c+id| \implies \sqrt{a^2+b^2} = \sqrt{c^2+d^2} \implies a^2+b^2 = c^2+d^2$.

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

We know that $|z^n| = |z|^n$. For $n=2$, $|z^2| = |z|^2$.

$|z+1| = |z+4-3| \le |z+4| + |-3| \le 3 + 3 = 6$.

If $x$ is a negative real number, it lies on the negative real axis. The principal argument is $\pi$.

This is a standard result check. Let's calculate: $|z| = \sqrt{1+4} = \sqrt{5}$. $z^2 = (1+2i)^2 = 1 - 4 + 4i = -3 + 4i$. $1-z^2 = 1 - (-3+4i) = 4 - 4i$. $7-z = 7 - (1+2i) = 6 - 2i$. $|f(z)| = \frac{|6-2i|}{|4-4i|} = \frac{\sqrt{36+4}}{\sqrt{16+16}} = \frac{\sqrt{40}}{\sqrt{32}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} = \frac{|z|}{2}$.

$|z - (-1+i)| = |z - (1-i)|$. This is the locus of a point equidistant from two fixed points $(-1, 1)$ and $(1, -1)$. This represents the perpendicular bisector of the segment joining these points, which is a straight line.

Given $|z - (-4)| \le 3 \implies |z+4| \le 3$. We need extrema of $|z+1|$. $|z+1| = |z+4-3|$. Max value: $|z+4| + 3 = 3+3 = 6$. Min value: $||z+4| - 3|$. Since center is at -4, distance to -1 is 3. The point -1 lies on the boundary of the circle $|z+4|=3$. So $z=-1$ is possible, giving $|-1+1|=0$. Thus, greatest is 6 and least is 0.