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If-Re-z-4-2z-1-1-2-then-z-is-represented-by-a-point-lying-on-1-A-circle-2-An-ellipse-3-A-straight-line-4-No-real-locus-




Question Number 20800 by Tinkutara last updated on 03/Sep/17
If Re(((z + 4)/(2z − 1))) = (1/2), then z is represented  by a point lying on  (1) A circle  (2) An ellipse  (3) A straight line  (4) No real locus
IfRe(z+42z1)=12,thenzisrepresentedbyapointlyingon(1)Acircle(2)Anellipse(3)Astraightline(4)Noreallocus
Commented by ajfour last updated on 03/Sep/17
Re(z)=(1/2) .  ⇒ straight line.
Re(z)=12.straightline.
Answered by ajfour last updated on 03/Sep/17
⇒       ((x+4+iy)/(2x−1+i2y))=(1/2)+iq  ⇒ (((x+4+iy)(2x−1−i2y))/((2x−1)^2 +4y^2 ))=(1/2)+iq  ⇒  (((x+4)(2x−1)+2y^2 )/((2x−1)^2 +4y^2 )) =(1/2)  or    ((2(x+4)(2x−1)+4y^2 )/((2x−1)^2 +4y^2 )) = 1  ⇒    2x−1=0   or  2(x+4)=2x−1  from first   x= Re(z)=(1/2)  from second  2x+8=2x−1                    no solution.       so  Re(z)=(1/2)     Im(z)≠0 .  straight line with one point  (z=1/2+0i)   removed.
x+4+iy2x1+i2y=12+iq(x+4+iy)(2x1i2y)(2x1)2+4y2=12+iq(x+4)(2x1)+2y2(2x1)2+4y2=12or2(x+4)(2x1)+4y2(2x1)2+4y2=12x1=0or2(x+4)=2x1fromfirstx=Re(z)=12fromsecond2x+8=2x1nosolution.soRe(z)=12Im(z)0.straightlinewithonepoint(z=1/2+0i)removed.
Commented by ajfour last updated on 03/Sep/17
do you have a solution?
doyouhaveasolution?
Commented by Tinkutara last updated on 03/Sep/17
Thank you very much Sir!
ThankyouverymuchSir!
Answered by Tinkutara last updated on 03/Sep/17
Commented by Tinkutara last updated on 03/Sep/17
This is in book.
Thisisinbook.
Commented by ajfour last updated on 03/Sep/17
thanks for posting.
thanksforposting.
Commented by Tinkutara last updated on 03/Sep/17

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