Prove that

Question:

If $(1+i) z=(1-i)^{\bar{z}}$ then prove that $z=-i \bar{z}$.

 

Solution:

Let z = x + iy

Then

$\bar{z}=x-i y$

Now, Given: $(1+i) z=(1-i) \bar{z}$

Therefore,

$(1+i)(x+i y)=(1-i)(x-i y)$

$x+i y+x i+i^{2} y=x-i y-x i+i^{2} y$

We know that $i^{2}=-1$, therefore,

$x+i y+i x-y=x-i y-i x-y$

$2 x i+2 y i=0$

$x=-y$

Now, as $x=-y$

$z=-\bar{Z}$

Hence, Proved.

 

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