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Beantwortet: Bestimmen Sie alle Lösungen der Form a+bi. (2z)/(1-i)+(16+2i)/(i-2)=Z(Quer)-7-4i

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(2z)/(1-i)+(16+2i)/(i-2)=Z(Quer)-7-4i

(2a + i2b)/(1-i)+(16+2i)/(i-2)=a -ib-7-4i

(2a + i2b)(1+i)/((1-i)(1+i)) +((16+2i)(i+2))/((i-2)(i+2)) =a -ib-7-4i

(2a + i2b + 2ai - b)/(1+1) +((16+2i)(i+2))/(-1 - 4) =a -ib-7-4i

(2a + i2b + 2ai - 2b)/2 +((16+2i)(i+2))/(-5) =a -ib-7-4i

a + ib + ai - b +(16i-2 + 32 + 4i)/(-5) =a -ib-7-4i

 i2b + ai - b +(20i + 30)/(-5) = -7-4i

 i2b + ai - b -(4i + 6) = -7-4i

 i2b + ai - b  = (4i + 6) -7-4i = -1 

i(2b + a) - b = - 1 + 0*i 

Realteil und Imaginärteil teilen

-b = - 1 ==> b = 1

2b + a = 0 ,

Weil b=1

==> 2 + a = 0 

==> a = -2

==> z = -2 + 1*i = -2 + i

Das war nun mal das Prinzip, wie man so was rechnen kann.

Unbedingt: Selbst rechnen und z.B. Vorzeichenfehler korrigieren. 


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