## Task B-3.1.

For complex numbers $z$ and $w$, it is given that $|z+w|=\sqrt{3}$ and $|z|=|w|=1$. Calculate $|z-w|$.
## First Solution.

Apply the property of the modulus of a complex number $|z|^{2}=z \cdot \bar{z}$ to the number $z+w$:

$$
|z+w|^{2}=(z+w) \cdot \overline{(z+w)}=(z+w)(\bar{z}+\bar{w})=z \cdot \bar{z}+w \cdot \bar{z}+z \cdot \bar{w}+w \cdot \bar{w}. \quad 1 \text { point }
$$

We have $z \cdot \bar{z}=|z|^{2}=1, w \cdot \bar{w}=|w|^{2}=1$ and $|z+w|^{2}=3$, 1 point

so $3=1+w \cdot \bar{z}+z \cdot \bar{w}+1$, which means $w \cdot \bar{z}+z \cdot \bar{w}=1$. 1 point

Then,

$$
\begin{aligned}
|z-w|^{2} & =(z-w) \cdot \overline{(z-w)}=(z-w)(\bar{z}-\bar{w}) \\
& =z \cdot \bar{z}-w \cdot \bar{z}-z \cdot \bar{w}+w \cdot \bar{w} \\
& =|z|^{2}-(w \cdot \bar{z}+z \cdot \bar{w})+|w|^{2} \\
& =1-1+1=1
\end{aligned}
$$

Therefore, $|z-w|=1$.