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Sektion 3.1

[Matematik/Kalkulus (Lindstroem)/Kapitel 3]

Opgave 1

Regn

a) $(2 + 3i) + (5 - 6i) = 7 - 3i$

b) $(4 + 8i) - (7 - 3i) = 4 - 7 + 8i - 3i = -3 + 5i$

c) $2i + 3(4 + i) = 2i +12 + 3i = 12 + 5i$

d) $(5 + 2i)(3 + i) = 15 + 5i + 6i + 2i^2 = 13 + 11i$

e) $(-2 + 3i)(2 - 2i) = -4 + 4i + 6i - 6i^2 = 10 + 10i$

f) $\frac{4 + 3i}{2 + i} = \frac{(2 - i)(4 + 3i)}{5} = \frac{8 + 6i - 4i - 3i^2}{5} = \frac{11}{5} + i\frac{2}{5}$

g)
$\frac{-5 + 2i}{5 - 5i} = \frac{(5 + 4i)(4 + 3i)}{25 + 16} = \frac{-25 + 10i - 20i + 8i^2}{41} = \frac{-33}{41} + i\frac{10}{41}$

h) $(5 - i)^{-1} = \frac{1}{5 - i} = \frac{5 + i}{25 + 1} = \frac{5}{26} + i\frac{1}{26}$

i) $(7 + 2i)^{-1} = \frac{1}{7 + 2i} = \frac{7 - 2i}{49 + 4} = \frac{7}{53} - i\frac{2}{53}$

Opgave 5

Løs ligningerne.

a) 2iz = 3 + 4i $$ 2iz = 3 + 4i \Rightarrow \\ 2i^{2}z = 3i - 4 \Rightarrow \\ z = \frac{-4}{-2} + i\frac{3}{-2} = 2 - i\frac{3}{-2} \\ $$

b) (1 + i)z + 3 = 1 - i $$ (1 + i)z + 3 = 1 - i \Rightarrow \\ (1 + i)z = -z + i \Rightarrow \\ z = \frac{-2 + i}{1 + i} \Rightarrow \\ z = \frac{(1 - i)(-2 + i)}{2} \Rightarrow \\ z = \frac{-2 + i + 2i - i^2}{2} \Rightarrow \\ z = \frac{-1}{2} + i\frac{3}{2} $$

c) (z - 2)/(z + 1) = 3i $$ \frac{z - 2}{x + 1} = 3i \Rightarrow \\ z - 2 = 3iz + 3i \Rightarrow \\ z - 3iz = 3i + 2 \Rightarrow \\ z(1 - 3i) = 3i + 2 \Rightarrow \\ z = \frac{3i + 2}{1 - 3i} = \frac{(1 + 3i)(3i + 2)}{10} \Rightarrow \\ z = \frac{3i + 2 + 9i^2 + 6i}{10} = \frac{-7}{10} + i\frac{9}{10} $$

d) (3 - 4i)/z = (2 + 3i)/(z - i) $$ \frac{3 - 4i}{z} = \frac{2 + 3i}{z - i} \Rightarrow \\ (3 - 4i)(z - i) = (2 + 3i)z \Rightarrow \\ 3z - 3i - 4iz + 4i^2 = 2z + 3iz \Rightarrow \\ z - 7iz = 4 + 3i \Rightarrow \\ z(1 + 7i) = 4 + 3i \Rightarrow \\ z = \frac{4 + 3i}{1 + 7i} \Rightarrow \\ z = \frac{(1 - 7i)(4 + 3i)}{50} = \frac{4 + 3i + 28i + 21i^2}{50} = \frac{-17}{50} + i\frac{31}{50} $$

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