If z = x + iy and |z - zi| = 1, then | vedclass

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$z$ lies on a circle

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(C) Given $|z - zi| = 1$. Substitute $z = x + iy$: $|x + iy - i(x + iy)| = 1$ $|x + iy - ix - i^2y| = 1$ Since $i^2 = -1$,we have: $|x + iy - ix + y| = 1$ $|(x + y) + i(y - x)| = 1$ Taking the modulus: $\sqrt{(x + y)^2 + (y - x)^2} = 1$ $(x + y)^2 + (y - x)^2 = 1$ $(x^2 + y^2 + 2xy) + (y^2 + x^2 - 2xy) = 1$ $2x^2 + 2y^2 = 1$ $x^2 + y^2 = \frac{1}{2}$ This is the equation of a circle with center $(0, 0)$ and radius $\frac{1}{\sqrt{2}}$. Thus,$z$ lies on a circle.

If $z = x + iy$ and $|z - zi| = 1,$ then

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