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1.1 Complex numbers and complex plane

1.

2.

(a) Let $z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$. Then

$\displaystyle \bar{z_{1} + z_{2}}$ $\displaystyle =$ $\displaystyle \bar{x_{1} + iy_{1} + x_{2} + iy_{2}} = x_{1} + x_{2} + i(y_{1} + y_{2})$  
  $\displaystyle =$ $\displaystyle x_{1} + x_{2} - i(y_{1} + y_{2}) = x_{1} - iy_{1} + x_{2} - iy_{2} = \bar{z_{1}} + \bar{z_{2}}$  

(b) Let $z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$. Then

$\displaystyle \bar{z_{1}z_{2}}$ $\displaystyle =$ $\displaystyle \bar{(x_{1} + iy_{1})(x_{2} + iy_{2})} = \bar{x_{1}x_{2} - y_{1}y_{2} + i(x_{1}y_{2} + x_{2}y_{1})}$  
  $\displaystyle =$ $\displaystyle x_{1}x_{2} - y_{1}y_{2} - i(x_{1}y_{2} + x_{2}y_{1}) = (x_{1} - iy_{1})(x_{2} - iy_{2})$  
  $\displaystyle =$ $\displaystyle \bar{z_{1}}\bar{z_{2}}$  

(c) Let $z = x + iy$. Then

$\displaystyle \frac{z + \bar{z}}{2}$ $\displaystyle =$ $\displaystyle \frac{x + iy + x - iy}{2} = x = Re{z}$  

3.

(a)

$\displaystyle \Re{z} = \frac{z + \bar{z}}{2} \leq \frac{\vert z\vert + \vert z\vert}{2} = \vert z\vert $

(b)

$\displaystyle \vert z_{1} + z_{2}\vert^2$ $\displaystyle =$ $\displaystyle (z_{1} + z_{2})\bar{z_{1} + z_{2}} = \vert z_{1}\vert^2 + z_{1}\bar{z_{2}} + \bar{z_{1}}z_{2} + \vert z_{2}\vert^2$  
  $\displaystyle =$ $\displaystyle \vert z_{1}\vert^2 + 2\Re{z_{1}\bar{z_{2}}} + \vert z_{2}\vert^2 \leq \vert z_{1}\vert^2 + 2\vert z_{1}\vert \vert z_{2}\vert + \vert z_{2}\vert^2$  
  $\displaystyle =$ $\displaystyle (\vert z_{1}\vert + \vert z_{2}\vert)^2$  

Take square root of both sides, we have $\vert z_{1} + z_{2}\vert \leq \vert z_{1}\vert + \vert z_{2}\vert$

(c) Since $Re{z} \leq \vert z\vert$, we note that $-\vert z\vert \leq -\Re{z}$.

$\displaystyle \vert\vert z_{1}\vert - \vert z_{2}\vert\vert^2 = \vert z_{1}\ver... ...2 - 2\Re{z_{1}\bar{z_{2}}} + \vert z_{2}\vert^2 \leq \vert z_{1} - z_{2}\vert^2$

Therefore, $\vert\vert z_{1}\vert - \vert z_{2}\vert\vert \leq \vert z_{1} - z_{2}\vert$

4.

(a) Cartesian coordinates to polar coordinates is given by $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

Write $-1 + i$ in polar coordinates, it is enough to find $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ and $\theta = \tan^{-1}{-1} = -\frac{\pi}{4}$. Thus,

$\displaystyle -1 + i = \sqrt{2}e^{-pi i/4} $

(b) Cartesian coordinates to polar coordinates is given by $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

Write $3 - \sqrt{3}i$ in polar coordinates, it is enough to find $r = \sqrt{(3)^2 + (\sqrt{3}^2} = \sqrt{12} = 2\sqrt{3}$ and $\theta = \tan^{-\sqrt{3}}{3} = -\frac{\pi}{6}$. Thus,

$\displaystyle 3 - \sqrt{3} i = 2\sqrt{3}e^{-pi i/6} $

(c) Cartesian coordinates to polar coordinates is given by $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

$\displaystyle -1 = -1 + 0i = e^{pi} $

(d) Cartesian coordinates to polar coordinates is given by $r = \sqrt{x^2 + y^2}$ , $\theta = \tan^{-1}{\frac{y}{x}}$ for $- \pi < \theta \leq pi$.

$\displaystyle 2i = 0 + 2i = 2e^{pi/2} $

5.

(a) $\arg{z}$ is the angle between the $x$ axis and the straight line drawn from the origin to the point $z$. Therefore, this is constant because the set of points $z$ is a point that forms a constant angle with the $x$ axis from the origin, so it is a straight line.

Alternate solution $\arg{z} =$   constant means that for some constant $c$, $\tan^{-1}{(y/x)} = c$. Thus $\frac{y}{x} = \tan{c}$ implies $y = \tan{c}x$. Therefore, it is a straight line emitted from the origin.

(b) $\vert z\vert =$   constant means that the distance from the origin is constant. Therefore, it is a circle.

(c) $\vert z-1\vert$ means that the distance from point 1. Also, $\vert z-i\vert$ is the distance from the point $i$. A collection of points where the two are equal is a perpendicular bisector passing through point 1 and point $i$.

Alternate solution $\vert z-1\vert = \sqrt{(x-1)^2 + y^2}$, $\vert z-i\vert = \sqrt{x^2 + (y-1)^2}$. Then rewrite $\vert z - 1\vert = \vert z - i\vert$.

$\displaystyle \sqrt{(x-1)^2 + y^2} = \sqrt{x^2 + (y-1)^2} $

Thus,

$\displaystyle (x-1)^2 + y^2 = x^2 + (y-1)^2 \Rightarrow 2x - 2y = 0 \Rightarrow y = x $

(d) $\vert z-2i\vert$ is the distance from point $2i$. Then $\vert z - 2i\vert = 3$ is the circle with the radius 3 centered at $2i$.

Alternate solutoin $\vert z-2i\vert = \sqrt{x^2 + (y-2)^2}$繧医ｊ Rewrite $\vert z - 2i\vert = 3$

$\displaystyle {x^2 + (y-2)^2} = 3^2$

Thus it is a circle with the radius 3 centered at $2i$.

(e) $\vert z+3\vert$ means $\vert z - (-3)\vert$. Then it is the distance from the point $-3$. $\vert z-1\vert$ is the distance from the point $1$. Thus $\vert z + 3\vert = 3\vert z - 1\vert$ means that the distance from the point $-3$ is 3 times the distance from the point $1$. Such a point draws a circle whose diameter is the point that internally divides the straight line connecting point-3 and point 1 into 3: 1 and the point that divides it outward. This circle is called Apollonius circle.

Alternate solution $\vert z+3\vert = \sqrt{(x+3)^2 + y^2}$, $3\vert z-1\vert = 3\sqrt{(x-1)^2 + y^2}$. Then rewrite $\vert z + 3\vert = 3\vert z - 1\vert$

$\displaystyle (x+3)^2 + y^2 = 9[(x-1)^2 + y^2]$

Simplify

$\displaystyle 8x^2 -18x + 8y^2 = 0 $

or,

$\displaystyle 8[(x - \frac{9}{8})^2 + y^2] = \frac{81}{8} $

Thus it is a circle with the radius $\frac{9}{2\sqrt{2}}$ centered at $\frac{9}{8}$.

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