Sei z ein komplexe Zahl. Zeigen Sie, dass (a) z + ¯z = 2Re(z) (b) z − ¯z = 2iIm(z) (c) z¯z = |z|2.

Question

7. Sei z ein komplexe Zahl. Zeigen Sie, dass
(a) z + ¯z = 2Re(z)
(b) z − ¯z = 2iIm(z)
(c) z¯z = |z|2.

Answer 1

Gezeigt werden soll:

(a) z + z* = 2Re(z)
(b) z − z* = 2iIm(z)
(c) z z* = |z|^2

 

Annahme:

z = a +bi;   Re{z} = a;   Im{z} = b;

z* = a -bi;  Re{z} = a;   Im{z} = -b;

|z| = sqrt(a^2 +b^2);

 

a) a +bi + a -bi = 2*a = 2*Re{z};

b) a +bi -a +bi = 2*b*i = 2*Im{z}*i;

c) (a +bi) *(a -bi) = a^2 -abi +abi +b^2 = a^2 +b^2 = |z|^2;

 

lg JR

Answer 2

(a)

$$ z=Re(z)+Im(z)\cdot i, \bar(z)=Re(z)-Im(z)\cdot i $$ $$ \Rightarrow z+\bar{z}=Re(z)+Im(z)\cdot i+Re(z)-Im(z)\cdot i=2Re(z) $$

(b)

$$ z=Re(z)+Im(z)\cdot i, \bar(z)=Re(z)-Im(z)\cdot i $$ $$ \Rightarrow z-\overline{z}=Re(z)+Im(z)\cdot i-(Re(z)-Im(z)\cdot i) $$ $$ =Re(z)+Im(z)\cdot i-Re(z)+Im(z)\cdot i=2Im(z)\cdot i $$

(c)

$$ z=Re(z)+Im(z)\cdot i, \bar(z)=Re(z)-Im(z)\cdot i, |z|=\sqrt{Re(z)^2+Im(z)^2} $$ $$ \Rightarrow z\cdot \bar{z}=(Re(z)+Im(z)\cdot i)\cdot (Re(z)-Im(z)\cdot i) $$ $$ =Re(z)^2-Re(z)Im(z)\cdot i+Re(z)Im(z)\cdot i-Im(z)^2\cdot i^2 $$ $$ = Re(z)^2-Im(z)^2\cdot (-1)=Re(z)^2+Im(z)^2=|z|^2 $$