Answer to Question #118181 in Algebra for carl

Question #118181
Find the Cartesian equation of the locus of the point P representing the complex
number z. Sketch the locus of P in each case.

(a) 2|z + 1| = |z − 2|

(b) |(z + i )/ z - 5 - 2i | = 1

(c) Im (z + (9/z ) ) = 0
1
Expert's answer
2020-05-27T18:48:38-0400

(a) 2z+1=z22|z + 1| = |z − 2|

Solution:

Let z=x+iyz=x+iy

Substitute z in (a)

2x+iy+1=x+iy22|x+iy+1|=|x+iy-2|

2x+iy+1=x+iy22\sqrt{\smash[b]{x+iy+1}}=\sqrt{\smash[b]{x+iy-2}}

2x+1+iy=x2+iy2\sqrt{\smash[b]{x+1+iy}}=\sqrt{\smash[b]{x-2+iy}}

2(x+1)2+y2=(x2)2+y22\sqrt{\smash[b]{(x+1)^2+y^2}}=\sqrt{\smash[b]{(x-2)^2+y^2}}

2((x+1)2+y2)=(x2)2+y22((x+1)^2+y^2)=(x-2)^2+y^2

2x2+4x+2+2y2=x24x+4+y22x^2+4x+2+2y^2=x^2-4x+4+y^2

2x2x2+2y2y2+4x+4x=422x^2-x^2+2y^2-y^2+4x+4x=4-2

Answer: x2+y2+8x=2x^2+y^2+8x=2




(b) (z+i)/z52i=1|(z + i )/ z - 5 - 2i | = 1

Solution:

Cross multiply equation to get;

z+i=z52i|z+i|=|z-5-2i|

Let z=x+iyz=x+iy

Substitute for z;

x+iy+i=x+iy52ix+iy+i=x+iy-5-2i

x+i(y+1)=x5+i(y2)x+i(y+1)=x-5+i(y-2)

x2+(y+1)2=(x5)2+(y2)2\sqrt{\smash[b]{x^2+(y+1)^2}}=\sqrt{\smash[b]{(x-5)^2+(y-2)^2}}

x2+y2+2y+1=x210x+25+y24y+4x^2+y^2+2y+1=x^2-10x+25+y^2-4y+4

Simplify;

x2x2+y2y2+10x+2y+4y=25+41x^2-x^2+y^2-y^2+10x+2y+4y=25+4-1

Answer : 10x+6y=2810x+6y=28




(c) Im(z+9z)=0Im (z + \frac{9}{z}) = 0

Solution:

Let z=x+iyz =x+iy

Substitute for z in the expression;

x+iy+9x+iyx+iy+\frac{9}{x+iy}

Simplify; x+iy(x+iy)x+iy+9x+iy\frac{x+iy(x+iy)}{x+iy}+\frac{9}{x+iy}

(x+iy)(x+iy)+9x+iy\frac{(x+iy)(x+iy)+9}{x+iy}

x2+2ixyy2+9x+iy\frac{x^2+2ixy-y^2+9}{x+iy}

Rationalize the denominator;

x2+2ixyy2+9x+iyxiyxiy\frac{x^2+2ixy-y^2+9}{x+iy}*\frac{x-iy}{x-iy}

x2(xiy)+2ixy(xiy)y2(xiy)+9(xiy)x2+y2\frac{x^2(x-iy)+2ixy(x-iy)-y^2(x-iy)+9(x-iy)}{x^2+y^2}

x3ix2y+2ix2y+2xy2y2x+iy3+9x9iyx2+y2\frac{x^3-ix^2y+2ix^2y+2xy^2-y^2x+iy^3+9x-9iy}{x^2+y^2}

Put the like terms together;

x3+2xy2y2x+9xix2y+2ix2y+iy39iyx2+y2\frac{x^3+2xy^2-y^2x+9x-ix^2y+2ix^2y+iy^3-9iy}{x^2+y^2}

x3+2xy2y2x+9xx2+y2+i(x2y+y39y)x2+y2\frac{x^3+2xy^2-y^2x+9x}{x^2+y^2} +\frac{i(x^2y+y^3-9y)}{x^2+y^2}

ImIm part states;

(x2y+y39y)x2+y2=0\frac{(x^2y+y^3-9y)}{x^2+y^2}=0

x2y+y39y=0x^2y+y^3-9y=0

y(x2+y29)=0y(x^2+y^2-9)=0

x2+y29=0x^2+y^2-9=0

Answer: x2+y2=9x^2+y^2=9


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