Question

find what curve zz' + (1+i)z + (1-i)z' =0 represents.​

Answer 1

Answer:

Given,

∣z−1∣=∣z+i∣

To find the locus of the equation.

Solution,

Given that,

∣z−1∣=∣z+i∣

Say x=x+iy

∣x+iy−1∣=∣x+iy+1∣

∣(x−1)+iy∣=∣x+i(y+1)∣

squiring both sides

2(x−1)(iy)+(x−1)

2

+y

2

i

2

=x

2

+i

2

(y+1)

2

+2xi(y+1)

(x−1)

2

−y

2

+2xyi−2yi=x

2

−y

2

−1−2y+2xyi+2xi

x

2

+1−2x−y

2

−2yi=x

2

−y

2

−2y+2xi−1

2y−2x+2=2xi+2yi

2(y−x+1)=2i(x+y)

y−x+1=i(x+y)=0

y−iy−x−xi+1=0

y(1−i)−x(1+i)+1=0

constantconstant

Hence It can be written as

by−ax+1=0

∴It represent the straight line.