If (1+i)z=(1-i)z conjugate then show that z=-iz conjugate
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Let z=a+ibz=a+ib. Your condition is equivalent
to a−iba+ib=1+i1−ia−iba+ib=1+i1−i.
But 1+i1−i=(1+i)22=i1+i1−i=(1+i)22=i.
Therefore, a−ib=i(a+ib)a−ib=i(a+ib), which is equivalent
to a−iba+ib=1+i1−ia−iba+ib=1+i1−i.
But 1+i1−i=(1+i)22=i1+i1−i=(1+i)22=i.
Therefore, a−ib=i(a+ib)a−ib=i(a+ib), which is equivalent
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