3.1 Übungen

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===Exercise 3.1:1===
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===Übung 3.1:1===
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Write in the form <math>\,a+bi\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers
Write in the form <math>\,a+bi\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers
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===Exercise 3.1:2===
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===Übung 3.1:2===
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Write in the form <math>\,a+bi\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers,
Write in the form <math>\,a+bi\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers,
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===Exercise 3.1:3===
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===Übung 3.1:3===
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Determine the real number <math>\,a\,</math> such that the expression <math>\ \displaystyle\frac{3+i}{2+ai}\ </math> becomes purely imaginary (i.e. the real part equals zero).
Determine the real number <math>\,a\,</math> such that the expression <math>\ \displaystyle\frac{3+i}{2+ai}\ </math> becomes purely imaginary (i.e. the real part equals zero).
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===Exercise 3.1:4===
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===Übung 3.1:4===
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Solve the equations
Solve the equations

Version vom 13:23, 10. Mär. 2009

       Theory          Übungen      

Übung 3.1:1

Write in the form \displaystyle \,a+bi\,, where \displaystyle \,a\, and \displaystyle \,b\, are real numbers

a) \displaystyle (5-2i)+(3+5i) b) \displaystyle 3i -(2-i)
c) \displaystyle i(2+3i) d) \displaystyle (3-2i)(7+5i)
e) \displaystyle (1+i)(2-i)^2 f) \displaystyle i^{\,20} + i^{\,11}

Übung 3.1:2

Write in the form \displaystyle \,a+bi\,, where \displaystyle \,a\, and \displaystyle \,b\, are real numbers,

a) \displaystyle \displaystyle\frac{3-2i}{1+i} b) \displaystyle \displaystyle\frac{3i}{4-6i} - \displaystyle\frac{1+i}{3+2i}
c) \displaystyle \displaystyle\frac{(2-i\sqrt{3}\,)^2}{1+i\sqrt{3}} d) \displaystyle \displaystyle\frac{5-\displaystyle\frac{1}{1+i}}{3i + \displaystyle\frac{i}{2-3i}}

Übung 3.1:3

Determine the real number \displaystyle \,a\, such that the expression \displaystyle \ \displaystyle\frac{3+i}{2+ai}\ becomes purely imaginary (i.e. the real part equals zero).


Übung 3.1:4

Solve the equations

a) \displaystyle z+3i=2z-2 b) \displaystyle (2-i) z= 3+2i
c) \displaystyle iz+2= 2z-3 d) \displaystyle (2+i) \overline{z} = 1+i
e) \displaystyle \displaystyle\frac{iz+1}{z+i} = 3+i f) \displaystyle (1+i)\overline{z}+iz = 3+5i