QMISMF: Chapter 2 « Dan's Thoughts

5Oct/090

QMISMF: Chapter 2

2-1:

> (+ 3-i 2+4i)
5+3i
> (+ 1+3i 2)
3+3i
> (- -5+2i 2+2i)
-7
> (+ -2+i 2+2i)
+3i
> (* 3-i 2+4i)
10+10i
> (* 1+3i 2)
2+6i
> (* 0+i 1+3i)
-3+i
> (* -5+2i 2+3i)
-16-11i
> (* 2+3i -2+3i)
-13
> (* 2+3i 3+2i)
+13i

2-2:

$z^{-1}=\frac{1}{x^2+y^2}(x+iy)$
$\left(\frac{1}{i}\right)^{-1} \rightarrow 0-i=\frac{1}{1}(i)=i$
$i \cdot \frac{1}{i}= 1$
$\left(\frac{1}{-i}\right)^{-1} \rightarrow \frac{-i}{-1} = i = \frac{1}{1}(-i)=-i$
$-i \cdot \frac{1}{-i}= 1$

2-3:

$i^*i = -i \cdot i = 1$
$(-i)^*(-i)= i \cdot -i = 1$
$|i|^2=(i^*i)=1$
$|-i|^2=\left( (-i)^*(-i) \right)=1$
$|i| = \sqrt{i^*i} = \sqrt{-i \cdot i} = 1$
$|-i| = \sqrt{(-i)^*(-i)}= \sqrt{i \cdot -i} = 1$

2-4:

$z=3+4i$
$z^* = 3-4i$
$z^*z = (3-4i)(3+4i) = 25$
$|z| = \sqrt{z^*z} = \sqrt{25} = 5$
$z^2 = z \cdot z = (3+4i)(3+4i) = -7+24i$
$\frac{1}{z} = \frac{1}{z} \cdot \frac{z^*}{z^*} = \frac{z^*}{zz^*} = \frac{3-4i}{25}$

2-5:

$|w+z| < |w| + |z|$
$|3+4i| < |3| + |4i|$
$5 < 7$

2-6:

$\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2}}{2}} \cdot \left( \sqrt{1 + \frac{x}{\sqrt{x^2+y^2}}} + i\sqrt{1 - \frac{x}{\sqrt{x^2+y^2}}} \right)$
for $y \ge 0$
$\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2}}{2}} \cdot \left( -\sqrt{1 + \frac{x}{\sqrt{x^2+y^2}}} + i\sqrt{1 - \frac{x}{\sqrt{x^2+y^2}}} \right)$
for $y \le 0$

$\sqrt{i} = \sqrt{0+1i} = \sqrt{\frac{\sqrt{0^2+1^2}}{2}} \cdot \left( \sqrt{1 + \frac{0}{\sqrt{0^2+1^2}}} + i\sqrt{1 - \frac{0}{\sqrt{0^2+1^2}}} \right)$
$\sqrt{i} = \sqrt{0+1i} = \sqrt{\frac{1}{2}} \cdot (1+i) = \sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}$
$\left(\sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}\right)^2 = \frac{1}{2} + i - \frac{1}{2} = i$

$\sqrt{-i} = \sqrt{0-1i} = \sqrt{\frac{\sqrt{0^2+(-1)^2}}{2}} \cdot \left( -\sqrt{1 + \frac{0}{\sqrt{0^2+(-1)^2}}} + i\sqrt{1 - \frac{0}{\sqrt{0^2+(-1)^2}}} \right)$
$\sqrt{-i} = \sqrt{0-1i} = \sqrt{\frac{1}{2}} \cdot (-1+i) = -\sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}$
$\left(- \sqrt{\frac{1}{2}}+i \sqrt{\frac{1}{2}}\right) \cdot \left(- \sqrt{\frac{1}{2}}+i \sqrt{\frac{1}{2}}\right)=\frac{1}{2}-i-\frac{1}{2}=-i$

2-7:

$z=-\frac{b}{2a} \pm \sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$
$z=-\frac{b}{2a} \pm \sqrt{\frac{b^2-4ac}{2^2a^2}}$
$z=-\frac{b}{2a} \pm \frac{1}{2a}\sqrt{b^2-4ac}$
$z=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ since the above is now in the usual form of the quadratic formula it is clear z is a solution.