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:

latex path not specified.: |w+z| < |w| + |z|[/latex] [latex]|3+4i| < |3| + |4i|[/latex] [latex]5 < 7[/latex]  <strong>2-6:</strong>   [latex]\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.

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