Question

find what curve zz' + (1+i)z + (1-i)z' =0 represents.(here z' is the conjugate of z)​

Answer 1

Step-by-step explanation:

Let $z= x+iy$

Then, $\bar{z}=x-iy$

Now,

$z \bar{z} + (1 + i)z + (1 - i) \bar{z} = 0$

$\implies \: | z | ^{2} + (1 + i)(x + iy)+ (1 - i) (x - iy)= 0$

$\implies \: x^{2} + {y}^{2} + (x + ix + iy + {i}^{2} y)+ (x - ix - iy + {i}^{2} y)= 0$

$\implies \: x^{2} + {y}^{2} + x + (x + y )i - y+ x - (x + y )i - y= 0$

$\implies \: x^{2} + {y}^{2} + x - y+ x - y= 0$

$\implies \: x^{2} + {y}^{2} + 2x - 2y= 0$

So, the required curve is a circle passing through origin, centre$\equiv(-1,1)$