UWIS, Mathematik 1, Lösung Serie 11

Thomas Kuster 26. Januar 2004

1 Bestimme Re z, Im z, |z|, arg z

1.1

|\frac{1}{z}| = 3 ⇒ |z| = \frac{1}{3}
arg z̄ = \frac{4π}{3} ⇒ arg z = \frac{2π}{3}
\frac{a}{\frac{1}{3}} = cos \frac{2π}{3}
3a = −\frac{1}{2}
a = −\frac{1}{6}
\sqrt{a^{2} + b^{2}} = \frac{1}{3}
(−\frac{1}{6})^{2} + b^{2} = \frac{1}{36} + b^{2} = \frac{1}{9}
b^{2} = \frac{1}{9} − \frac{1}{36} = \frac{4 − 1}{36}
b = \sqrt{\frac{3}{36}} = \sqrt{\frac{1}{12}}
Re z = −\frac{1}{6}
Im z = \frac{1}{\sqrt{12}}

UWIS, Mathematik 1, Lösung Serie 11 2

1.2

z = \frac{−1 + i\sqrt{3}}{1 + i\sqrt{3}} = \frac{(−1 + i\sqrt{3})(1 − i\sqrt{3})}{(1 + i\sqrt{3})(1 − i\sqrt{3})} = \frac{−1 + 2i\sqrt{3} − 3i^{2}}{1 − 3i^{2}}
z = \frac{2 + 2i\sqrt{3}}{4} = \frac{1}{2} + \frac{\sqrt{3}}{2}i
Re z = \frac{1}{2}
Im z = \frac{\sqrt{3}}{2}
|z| = \sqrt{(\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} = 1
arg z = arctan \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = arctan \sqrt{3} = \frac{π}{3}

1.3

Re (iz̄) = Re (i(a − ib)) = Re (ai − bi^{2}) = Re ai + b = 3
b = 3
|iz̄| = |i(a − ib)| = |ia + b| = \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + 3^{2}} = 5
a^{2} + 9 = 25
a^{2} = 16
a = ±4
Re z = ±4
Im z = 3
|z| = 5
arg z = ± arctan \frac{3}{4}

UWIS, Mathematik 1, Lösung Serie 11 3

1.4

2 = Im \frac{z}{i} = Im \frac{a + ib}{i} = Im \frac{(a + ib)i}{i^{2}} = Im \frac{ia − b}{−1} = Im (b − ia)
a = −2
−\frac{π}{3} = arg (iz̄) = arg i(a + ib) = arg (ia − b) = arg (−2i − b)
\frac{π}{3} = arg (−b + 2i)
\frac{2}{−b} = tan \frac{π}{3} = \sqrt{3}
b = −\frac{2}{\sqrt{3}}
Re z = −2
Im z = −\frac{2}{\sqrt{3}}
|z| = \sqrt{(−2)^{2} + (−\frac{2}{\sqrt{3}})^{2}} = \frac{4}{\sqrt{3}}
arg z = arctan \frac{−\frac{2}{\sqrt{3}}}{−2} = arctan \frac{1}{\sqrt{3}} = \frac{π}{6}

2 Schreibe in der Form z = a + ib

2.1

2.1.1 z_{1}

\frac{\sqrt{3} + 1 − i (\sqrt{3} + 1)}{1 + i} = \frac{(\sqrt{3} + 1 − i\sqrt{3} − i) (1 − i)}{(1 + i)(1 − i)}

= \frac{\sqrt{3} + 1 − i\sqrt{3} − i − i\sqrt{3} − i − i^{2}\sqrt{3} − i^{2}}{1 − i^{2}} =\frac{2\sqrt{3} + 2 − 2\sqrt{3}i − 2i}{2}
=\sqrt{3} + 1 + i(−\sqrt{3} − 1)

2.1.2 z_{2}

\begin{matrix}(\sqrt{3} + i)^{6} (1 − i) = (\sqrt{\sqrt{3}^{2} + 1^{2}} e^{arctan^{( \frac{1}{\sqrt{3}} )}})^{6} (1 − i) \\ = 2^{6} \underbrace{e^{\frac{π}{6} 6}}_{=−1}(1 − i) = −64(1 − i) = −64 + i64 \end{matrix}

2.1.3 z_{3} e^{−i \frac{π}{4}} ⇒ gleichschenkliges (a = b), rechtwinkliges Dreieck wobei die Hypotenuse die Länge 1 hat.

1^{2} = a^{2} + a^{2}
\sqrt{\frac{1}{2}} = a
a + ib = \frac{1}{\sqrt{2}} − i\frac{1}{\sqrt{2}}

2.1.4 z_{4}

\frac{1}{e^{i \frac{2π}{3}}} = e^{−i \frac{2π}{3}} = e^{i \frac{4π}{3}}
Im z = sin\frac{4π}{3} = −\frac{\sqrt{3}}{2}
Re z = cos\frac{4π}{3} = −\frac{1}{2}
a + ib = −\frac{\sqrt{3}}{2} − i\frac{1}{2}

2.1.5 z_{5}

i \underbrace{e^{−i \frac{π}{6}}}_{z_{1}}

Im z_{∗} = sin−\frac{π}{6} = −\frac{1}{2}
Re z_{∗} = cos−\frac{π}{6} = \frac{\sqrt{3}}{2}
i (\frac{\sqrt{3}}{2} − i\frac{1}{2}) = i\frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}

2.1.6 z_{6}

i^{i} = (e^{i \frac{π}{2}} )^{i} = e^{i^{2} \frac{π}{2}} = e^{− \frac{π}{2}} ≈ 0.209

2.2

2.2.1

arg z_{1} = arctan (\frac{−\sqrt{3} − 1}{\sqrt{3} + 1}) = arctan −1 = \frac{π}{4}

2.2.2

\begin{matrix}z_{3}^{2}z_{4} = (\frac{1}{\sqrt{2}} − i\frac{1}{\sqrt{2}})^{2} (−\frac{\sqrt{3}}{2} − i\frac{1}{2}) = (\frac{1}{2} − 2i\frac{1}{2} + i^{2} \frac{1}{2}) (−\frac{\sqrt{3}}{2} − i\frac{1}{2}) \\ = −i (−\frac{\sqrt{3}}{2} − i\frac{1}{2}) = i\frac{\sqrt{3}}{2} + i^{2} \frac{1}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2} \end{matrix}

2.2.3

\begin{aligned}\frac{z_{4}}{z_{5}} & = \frac{−\frac{\sqrt{3}}{2} − i\frac{1}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = −\begin{matrix}(\frac{\sqrt{3}}{2} + i\frac{1}{2}) & (\frac{1}{2} − i\frac{\sqrt{3}}{2}) \\ (\frac{1}{2} + i\frac{\sqrt{3}}{2}) & (\frac{1}{2} − i\frac{\sqrt{3}}{2}) \end{matrix} \\ & = \frac{\frac{\sqrt{3}}{4} − i\frac{3}{4} + i\frac{1}{4} + \frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{3}{4}} = \frac{2\sqrt{3}}{4} − i\frac{2}{4} = \frac{\sqrt{3}}{2} − i\frac{1}{2} \end{aligned}

3 Bestimme alle komplexen Zahlen z

3.1

Polynom zweiten Grades ⇒ 2 zwei Lösungen.

z^{2} = (a + ib)^{2} = i
a^{2} + 2iab − b^{2} = i
a^{2} − b^{2} = 0 ⇒ a = ±b
2iab = i
ab = \frac{1}{2} da a = ±b ⇒ a = b
a^{2} = \frac{1}{2}
a = b = ±\frac{1}{\sqrt{2}}

Lösungen: \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} und −\frac{1}{\sqrt{2}} − i\frac{1}{\sqrt{2}}

3.2

Eine Lösung von z^{3} = −1 kann erraten werden: (−1)^{3} = −1.Die anderen drei Lösungen sind gleichmässig auf dem Einheitskreis verteilt da −1 = e^{iπ} ist sind die zwei Lösungen: e^{i \frac{π}{3}} und e^{i \frac{2π}{3}} .

z_{1} = −1
z_{2} = cos \frac{π}{3} + i sin \frac{π}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}
z_{3} = \frac{1}{2} − i\frac{\sqrt{3}}{2}

UWIS, Mathematik 1, Lösung Serie 11 6

3.3

\begin{matrix}z^{4} = & \frac{81 (\sqrt{3} + i)}{−\sqrt{3} + i} = \frac{81 (\sqrt{3} + i) (−\sqrt{3} − i)}{(−\sqrt{3} + i) (−\sqrt{3} − i)} = \frac{81 (−3 − 2i\sqrt{3} + 1)}{3 + 1} \\ = & \frac{81 · −2 (1 + i\sqrt{3})}{4} = −\frac{81}{2} − \frac{i81\sqrt{3}}{2} \end{matrix}

r = \sqrt{\frac{81^{2}}{4} + \frac{3 · 81^{2}}{4}} = \sqrt{\frac{81^{2}(1 + 3)}{4}} = 81
φ = arctan \frac{−\frac{81\sqrt{3}}{2}}{−\frac{81}{2}} = arctan \sqrt{3} = \frac{π}{3}
z^{4} = 81e^{i(\frac{π}{3} +2kπ)}
z = 3e^{i(\frac{π}{3·4} + \frac{2kπ}{4})} = 3e^{i(\frac{π}{12} + \frac{6kπ}{12})}
z_{1} = 3e^{i \frac{π}{12}}
z_{2} = 3e^{i \frac{7π}{12}}
z_{3} = 3e^{i \frac{13π}{12}}
z_{4} = 3e^{i \frac{19π}{12}}

3.4

4zz̄ + (z − z̄)^{2} = 3
4(a + ib)(a − ib) + ((a + ib) − (a − ib))^{2} = 3
4(a^{2} + b^{2})(2ib)^{2} = 3
4a^{2} + 4b^{2} + 4i^{2}b^{2} = 3
4a^{2} + 4b^{2} − 4b^{2} = 3
4a^{2} = 3
a = ±\frac{\sqrt{3}}{2}
b = ∈ ℝ

3.5

iz^{2} + 2z + \sqrt{3} = 0

z_{1} = \frac{−2 + \sqrt{2^{2} − 4i\sqrt{3}}}{2i}

z_{2} = \frac{−2 − \sqrt{2^{2} − 4i\sqrt{3}}}{2i}

UWIS, Mathematik 1, Lösung Serie 11 7

4 Teich mit Forellen

4.1

V_{0} = 17280lzur Zeit t_{0}
N(t) = Anzahl Forellen im Teich zur Zeit t_{0}
N(0) = 1864 Forellen zur Zeit t_{0}
z = 12\frac{l}{min}
a = 8\frac{l}{min}
z_{F} = \frac{2}{40} \frac{Forellen}{l}
k = 12\frac{2}{40} = \frac{3}{5} \frac{Forellen}{min} Zufluss Forellen
α = Proportionalitätsfaktor

Wassermenge im Teich ist egal, da die Anzahl abfliessenden Forellen proportinal zur Anzahl Forellen im Teich ist.

N(0 + Δt) = N(0) + Δtk − ΔtN(0)α
\frac{N(0 + Δt) − N(0)}{Δt} = k − N(0)α
Ṅ(t) = k − N(t)α
N(t) = ce^{−αt} + k
N(0) = 1864 = c \underbrace{e^{−α0}}_{=1} +\frac{3}{5}
1864 = c + \frac{3}{5}
1864 − \frac{3}{5} = c
c = \frac{9317}{5} = 1863.4
N(t) = 1863.4e^{−αt} + \frac{3}{5}

4.2