3.3 Übungen
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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|width="50%"| <math>\displaystyle\frac{(1+i\sqrt{3}\,)(1-i)^8}{(\sqrt{3}-i)^9}</math> | |width="50%"| <math>\displaystyle\frac{(1+i\sqrt{3}\,)(1-i)^8}{(\sqrt{3}-i)^9}</math> | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:1|Solution a|Solution 3.3:1a|Solution b|Solution 3.3:1b|Solution c|Solution 3.3:1c|Solution d|Solution 3.3:1d|Solution e|Solution 3.3:1e}} |
===Übung 3.3:2=== | ===Übung 3.3:2=== | ||
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|width="33%"| <math>\displaystyle\Bigl(\frac{z+i}{z-i}\Bigr)^2 = -1</math> | |width="33%"| <math>\displaystyle\Bigl(\frac{z+i}{z-i}\Bigr)^2 = -1</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:2|Solution a|Solution 3.3:2a|Solution b|Solution 3.3:2b|Solution c|Solution 3.3:2c|Solution d|Solution 3.3:2d|Solution e|Solution 3.3:2e}} |
===Übung 3.3:3=== | ===Übung 3.3:3=== | ||
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|width="50%"| <math>iz^2+(2+3i)z-1</math> | |width="50%"| <math>iz^2+(2+3i)z-1</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:3|Solution a|Solution 3.3:3a|Solution b|Solution 3.3:3b|Solution c|Solution 3.3:3c|Solution d|Solution 3.3:3d}} |
===Übung 3.3:4=== | ===Übung 3.3:4=== | ||
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|width="50%"| <math>\displaystyle\frac{1}{z} + z = \frac{1}{2}</math> | |width="50%"| <math>\displaystyle\frac{1}{z} + z = \frac{1}{2}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:4|Solution a|Solution 3.3:4a|Solution b|Solution 3.3:4b|Solution c|Solution 3.3:4c|Solution d|Solution 3.3:4d}} |
===Übung 3.3:5=== | ===Übung 3.3:5=== | ||
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|width="50%"| <math>(4+i)z^2+(1-21i)z=17</math> | |width="50%"| <math>(4+i)z^2+(1-21i)z=17</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:5|Solution a|Solution 3.3:5a|Solution b|Solution 3.3:5b|Solution c|Solution 3.3:5c|Solution d|Solution 3.3:5d}} |
===Übung 3.3:6=== | ===Übung 3.3:6=== | ||
<div class="ovning"> | <div class="ovning"> | ||
Determine the solution to <math>\,z^2=1+i\,</math> both in polar form and in the form <math>\,a+ib\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers. Use the result to calculate <math>\; \tan \frac{\pi}{8}\,</math>. | Determine the solution to <math>\,z^2=1+i\,</math> both in polar form and in the form <math>\,a+ib\,</math>, where <math>\,a\,</math> and <math>\,b\,</math> are real numbers. Use the result to calculate <math>\; \tan \frac{\pi}{8}\,</math>. | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:6|Solution|Solution 3.3:6}} |
Version vom 13:31, 10. Mär. 2009
| Theorie | Übungen |
Übung 3.3:1
Write the following number in the form \displaystyle \,a+ib\,, where \displaystyle \,a\, and \displaystyle \,b\, are real numbers:
| a) | \displaystyle (i+1)^{12} | b) | \displaystyle \displaystyle\Bigl(\frac{1+i\sqrt{3}}{2}\,\Bigr)^{12} |
| c) | \displaystyle (4\sqrt{3} -4i)^{22} | d) | \displaystyle \Bigl(\displaystyle\frac{1+i\sqrt{3}}{1+i}\,\Bigr)^{12} |
| e) | \displaystyle \displaystyle\frac{(1+i\sqrt{3}\,)(1-i)^8}{(\sqrt{3}-i)^9} |
Antwort
Laden...
Solution a
Solution b
Solution c
Solution d
Solution e
Übung 3.3:2
Solve the equations
| a) | \displaystyle z^4=1 | b) | \displaystyle z^3=-1 | c) | \displaystyle z^5=-1-i |
| d) | \displaystyle (z-1)^4+4=0 | e) | \displaystyle \displaystyle\Bigl(\frac{z+i}{z-i}\Bigr)^2 = -1 |
Antwort
Solution a
Solution b
Solution c
Solution d
Solution e
Übung 3.3:3
Complete the square of the following expressions
| a) | \displaystyle z^2 +2z+3 | b) | \displaystyle z^2 +3iz-\frac{1}{4} |
| c) | \displaystyle -z^2-2iz +4z+1 | d) | \displaystyle iz^2+(2+3i)z-1 |
Antwort
Solution a
Solution b
Solution c
Solution d
Übung 3.3:4
Solve the equations
| a) | \displaystyle z^2=i | b) | \displaystyle z^2-4z+5=0 |
| c) | \displaystyle -z^2+2z+3=0 | d) | \displaystyle \displaystyle\frac{1}{z} + z = \frac{1}{2} |
Antwort
Solution a
Solution b
Solution c
Solution d
Übung 3.3:5
Solve the equations
| a) | \displaystyle z^2-2(1+i)z+2i-1=0 | b) | \displaystyle z^2-(2-i)z+(3-i)=0 |
| c) | \displaystyle z^2-(1+3i)z-4+3i=0 | d) | \displaystyle (4+i)z^2+(1-21i)z=17 |
Antwort
Solution a
Solution b
Solution c
Solution d
Übung 3.3:6
Determine the solution to \displaystyle \,z^2=1+i\, both in polar form and in the form \displaystyle \,a+ib\,, where \displaystyle \,a\, and \displaystyle \,b\, are real numbers. Use the result to calculate \displaystyle \; \tan \frac{\pi}{8}\,.
Antwort
Solution
