Let $z$ and $w$ be two complex numbers such that $|z| \leq 1$, $|w| \leq 1$ and $|z + iw| = |z - i\bar{w}| = 2$. Then $z$ equals
Step-by-Step Solution
Key Concept: The triangle inequality forces $|z|=|w|=1$. Subtracting the two squared-modulus conditions gives $bc=0$, splitting into tractable cases. The solutions are $z=\pm1$.
**Step 1: Use triangle inequality bound**
$|z+iw| \leq |z|+|iw| \leq 1+1=2$. Reaching $2$ requires $|z|=|w|=1$.
**Step 2: Set up algebraic equations**
Let $z=a+bi$, $w=c+di$. From $|z+iw|^2=|(a-d)+(b+c)i|^2=4$ and $|z-i\bar{w}|^2=|(a-d)+(b-c)i|^2=4$. Subtracting: $(b+c)^2-(b-c)^2=0 \Rightarrow 4bc=0$, so $b=0$ or $c=0$.
**Step 3: Solve for b=0 (z real)**
With $b=0$: $(a-d)^2+c^2=4$. Using $|z|=1\Rightarrow|a|=1$ and $c^2+d^2\leq1$: solving gives $(a,d,c)=(1,-1,0)\Rightarrow z=1,w=-i$, or $(a,d,c)=(-1,1,0)\Rightarrow z=-1,w=i$.
**Step 4: Verify both solutions**
$z=1,w=-i$: $|1+i(-i)|=|2|=2$ ✓, $|1-i(i)|=|2|=2$ ✓. $z=-1,w=i$: $|-1+i(i)|=|-2|=2$ ✓, $|-1-i(-i)|=|-2|=2$ ✓.
Correct Answer: 3