Chapter 4 - Complex Numbers and Quadratic Equations

$(1 - i)^n (1 - \frac{1}{i})^n = (1 - i)^n (1 - (-i))^n = (1 - i)^n (1 + i)^n = ((1 - i)(1 + i))^n = (1 - i^2)^n = (1 - (-1))^n = 2^n$.

$\sum_{n=1}^{13} (i^n + i^{n+1}) = \sum_{n=1}^{13} i^n(1 + i) = (1 + i) \sum_{n=1}^{13} i^n$.
We know that the sum of any four consecutive powers of $i$ is zero.
$\sum_{n=1}^{13} i^n = (i^1 + i^2 + i^3 + i^4) + (i^5 + \dots + i^8) + (i^9 + \dots + i^{12}) + i^{13} = 0 + 0 + 0 + i = i$.
Therefore, the value is $(1 + i)(i) = i + i^2 = i - 1$.

$\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2+1^2} = \frac{2i}{2} = i$.
$\frac{1-i}{1+i} = \frac{(1-i)^2}{1^2+1^2} = \frac{-2i}{2} = -i$.
Expression becomes $(i)^3 - (-i)^3 = -i - (i) = -2i$.
So $x + iy = 0 - 2i$. Hence $(x, y) = (0, -2)$.

$\frac{(1+i)^2}{2-i} = \frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{4+1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + i\frac{4}{5}$.
Here $x = -\frac{2}{5}$ and $y = \frac{4}{5}$.
$x + y = -\frac{2}{5} + \frac{4}{5} = \frac{2}{5}$.

$\frac{1-i}{1+i} = -i$.
$(-i)^{100} = (-1)^{100} i^{100} = 1 \cdot (i^4)^{25} = 1$.
$a + ib = 1 + 0i$. Hence $(a, b) = (1, 0)$.

$\frac{1+a}{1-a} = \frac{1 + \cos \theta + i \sin \theta}{1 - \cos \theta - i \sin \theta} = \frac{2\cos^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2\sin^2 \frac{\theta}{2} - 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$
$= \frac{2\cos \frac{\theta}{2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})}{2\sin \frac{\theta}{2} (\sin \frac{\theta}{2} - i \cos \frac{\theta}{2})}$
$= \cot \frac{\theta}{2} \frac{\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}}{-i(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})} = \frac{\cot \frac{\theta}{2}}{-i} = i \cot \frac{\theta}{2}$.

Given $(1 + i)z = (1 - i)\bar{z}$.
$\frac{z}{\bar{z}} = \frac{1-i}{1+i} = \frac{(1-i)^2}{2} = \frac{-2i}{2} = -i$.
$\therefore z = -i\bar{z}$.

$z\bar{z} = |z|^2 = x^2 + y^2$.
$z + \bar{z} = 2x$.
The equation becomes $x^2 + y^2 + 2(2x) + b = 0 \implies x^2 + y^2 + 4x + b = 0$.
This represents a circle with centre $(-2, 0)$ and radius $\sqrt{4-b}$ (provided $b < 4$).

Let $z = x + iy$.
$\frac{z+2}{z-1} = \frac{(x+2)+iy}{(x-1)+iy} = \frac{((x+2)+iy)((x-1)-iy)}{(x-1)^2+y^2}$.
Real part = $\frac{(x+2)(x-1) + y^2}{(x-1)^2+y^2} = 4$.
$x^2 + x - 2 + y^2 = 4(x^2 - 2x + 1 + y^2)$.
$x^2 + y^2 + x - 2 = 4x^2 + 4y^2 - 8x + 4$.
$3x^2 + 3y^2 - 9x + 6 = 0 \implies x^2 + y^2 - 3x + 2 = 0$.
This represents a circle.

Let $z = x + iy$.
$\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} = \frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2+y^2} = \frac{x^2-1+y^2 + i(y(x+1)-y(x-1))}{(x+1)^2+y^2}$.
$\arg(\frac{z-1}{z+1}) = \frac{\pi}{4} \implies \tan^{-1}(\frac{2y}{x^2+y^2-1}) = 1$.
$\frac{2y}{x^2+y^2-1} = 1 \implies x^2 + y^2 - 2y - 1 = 0$.
This represents a circle.

Let $z = x + iy$. Then $|z| = \sqrt{x^2+y^2}$.
Equation: $\sqrt{x^2+y^2} = (x+1) + i(y+2)$.
Comparing imaginary parts: $0 = y + 2 \implies y = -2$.
Comparing real parts: $\sqrt{x^2+4} = x + 1$.
Squaring: $x^2 + 4 = x^2 + 2x + 1 \implies 2x = 3 \implies x = \frac{3}{2}$.
So $z = \frac{3}{2} - 2i$.

$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (a^2 + b^2)(|z_1|^2 + |z_2|^2)$.

$\sqrt{-25} \times \sqrt{-9} = (5i)(3i) = 15i^2 = -15$.

$\frac{(1-i)^3}{1-i^3} = \frac{1 - 3i + 3i^2 - i^3}{1 - (-i)} = \frac{1 - 3i - 3 + i}{1 + i} = \frac{-2 - 2i}{1 + i} = \frac{-2(1 + i)}{1 + i} = -2$.

Sum of any 4 consecutive powers of $i$ is 0. Since 1000 is a multiple of 4, the sum is 0.

Inverse = $\frac{1}{1+i} = \frac{1-i}{1^2+1^2} = \frac{1-i}{2}$.

If $z_1 + z_2$ is real, then the imaginary part of $z_1 + z_2$ is 0. This implies $\text{Im}(z_2) = -\text{Im}(z_1)$. In the context of standard properties, $z_2$ is often taken as the conjugate $\bar{z}_1$ (though strictly real parts can differ).

$\arg(\bar{z}) = -\arg(z)$. Thus sum is 0.

$|z+1| = |z+4-3|$. Max value is $3+3=6$. Min value is $3-3=0$ (since $z=-1$ lies on the boundary of $|z+4|\le 3$).

The equation $\arg(\frac{z-z_1}{z-z_2}) = \alpha$ represents a segment of a circle passing through $z_1$ and $z_2$.

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

Complex numbers cannot be ordered.

Multiplication by $-i$ rotates the point by $-90^\circ$ (clockwise).

By triangle inequality, $|z| + |z-1| \ge |z - (z-1)| = 1$.

It is the perpendicular bisector of the segment joining $1$ and $i$.

If $\text{Re}(z) = 0$, $z = iy$. $z^2 = -y^2$, which is purely real. So $\text{Im}(z^2) = 0$.

The perpendicular bisector of 2 and 4 is $x=3$. Points closer to 4 satisfy $x > 3$.

The condition implies $\arg(z_1) = \arg(z_2)$. It does not imply $\arg(z_1 - z_2) = 0$.

2 is a complex number with imaginary part 0.