Question Number 114840 by zakirullah last updated on 21/Sep/20

Answered by Aziztisffola last updated on 21/Sep/20

 (i) z=x+iy   z+z^− =x+iy+x−iy=2x=2Re(z)  (ii) z−z^− =x+iy−x+iy=2iy=2iIm(z)  (iii) zz^− =(x+iy)(x−iy)=x^2 −(iy)^2                 =x^2 −i^2 y^2 =x^2 +y^2                 =Re(z)^2 +Im(z)^2   (iv) z=z^− ⇒x+iy=x−iy  ⇒2iy=0⇒y=0 ⇒ z=x (real)  (v) z^− =−z ⇒x−iy=−x−iy  ⇒2x=0⇒x=0 ⇒z pure imaginary.

Commented byzakirullah last updated on 21/Sep/20

Sir alot of thanks.