Complex Numbers
Modulus Condition and Locus
Complex Numbers_PYQ
Grade 11
Question:
If $z=x+iy$ and $w=\dfrac{1-iz}{z-i}$, then $|w|=1$ implies that, in the complex plane
$z$ lies on the imaginary axis
$z$ lies on the real axis
$z$ lies on the unit circle
None of these
Step-by-Step Solution
Key Concept: $|1-iz|=|z-i|$: note $1-iz = -i(z - i) \cdot i = ...$; equivalently, $z$ is equidistant from $i$ and $-i$, whose perpendicular bisector is the real axis.
**Step 1: Translate |w|=1**
$|w|=1 \Rightarrow |1-iz|=|z-i|$.
**Step 2: Expand with z=x+iy**
$iz=-y+ix$, so $1-iz=(1+y)-ix$ and $z-i=x+(y-1)i$. Then $|1-iz|^2=(1+y)^2+x^2$ and $|z-i|^2=x^2+(y-1)^2$.
**Step 3: Solve**
$(1+y)^2=(y-1)^2 \Rightarrow 4y=0 \Rightarrow y=0$. So $z$ lies on the real axis.
Correct Answer: 2