Determine complex number(s) z such that z- i , iz- i and z-iz have the same modulus
∣z−i∣=∣iz−i∣=∣z−iz∣|z-i|=|iz-i|=|z-iz|∣z−i∣=∣iz−i∣=∣z−iz∣
z=x+iyz=x+iyz=x+iy
z−i=x+i(y−1)z-i=x+i(y-1)z−i=x+i(y−1)
iz−i=−y+i(x−1)iz-i=-y+i(x-1)iz−i=−y+i(x−1)
z−iz=(x+y)+i(y−x)z-iz=(x+y)+i(y-x)z−iz=(x+y)+i(y−x)
∣z−i∣=x2+(y−1)2|z-i|=\sqrt{x^2+(y-1)^2}∣z−i∣=x2+(y−1)2
∣iz−i∣=y2+(x−1)2|iz-i|=\sqrt{y^2+(x-1)^2}∣iz−i∣=y2+(x−1)2
∣z−iz∣=(x+y)2+(y−x)2|z-iz|=\sqrt{(x+y)^2+(y-x)^2}∣z−iz∣=(x+y)2+(y−x)2
x2+(y−1)2=y2+(x−1)2x^2+(y-1)^2=y^2+(x-1)^2x2+(y−1)2=y2+(x−1)2
−2y=−2x-2y=-2x−2y=−2x
x=yx=yx=y
y2+(x−1)2=(x+y)2+(y−x)2y^2+(x-1)^2=(x+y)^2+(y-x)^2y2+(x−1)2=(x+y)2+(y−x)2
2x2−2x+1=2x22x^2-2x+1=2x^22x2−2x+1=2x2
x=y=1/2x=y=1/2x=y=1/2
z=0.5+0.5iz=0.5+0.5iz=0.5+0.5i
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