(a) 2∣z+1∣=∣z−2∣
Solution:
Let z=x+iy
Substitute z in (a)
2∣x+iy+1∣=∣x+iy−2∣
2x+iy+1=x+iy−2
2x+1+iy=x−2+iy
2(x+1)2+y2=(x−2)2+y2
2((x+1)2+y2)=(x−2)2+y2
2x2+4x+2+2y2=x2−4x+4+y2
2x2−x2+2y2−y2+4x+4x=4−2
Answer: x2+y2+8x=2
(b) ∣(z+i)/z−5−2i∣=1
Solution:
Cross multiply equation to get;
∣z+i∣=∣z−5−2i∣
Let z=x+iy
Substitute for z;
x+iy+i=x+iy−5−2i
x+i(y+1)=x−5+i(y−2)
x2+(y+1)2=(x−5)2+(y−2)2
x2+y2+2y+1=x2−10x+25+y2−4y+4
Simplify;
x2−x2+y2−y2+10x+2y+4y=25+4−1
Answer : 10x+6y=28
(c) Im(z+z9)=0
Solution:
Let z=x+iy
Substitute for z in the expression;
x+iy+x+iy9
Simplify; x+iyx+iy(x+iy)+x+iy9
x+iy(x+iy)(x+iy)+9
x+iyx2+2ixy−y2+9
Rationalize the denominator;
x+iyx2+2ixy−y2+9∗x−iyx−iy
x2+y2x2(x−iy)+2ixy(x−iy)−y2(x−iy)+9(x−iy)
x2+y2x3−ix2y+2ix2y+2xy2−y2x+iy3+9x−9iy
Put the like terms together;
x2+y2x3+2xy2−y2x+9x−ix2y+2ix2y+iy3−9iy
x2+y2x3+2xy2−y2x+9x+x2+y2i(x2y+y3−9y)
Im part states;
x2+y2(x2y+y3−9y)=0
x2y+y3−9y=0
y(x2+y2−9)=0
x2+y2−9=0
Answer: x2+y2=9
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