Let C denote the set of all complex numbers. | vedclass

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(C) Given $A = \{(z, w) \mid |z| = |w|\}$ and $B = \{(z, w) \mid z^2 = w^2\}$.For any $(z, w) \in B$, we have $z^2 = w^2$, which implies $z^2 - w^2 =

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$B \subset A$

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(C) Given $A = \{(z, w) \mid |z| = |w|\}$ and $B = \{(z, w) \mid z^2 = w^2\}$.For any $(z, w) \in B$, we have $z^2 = w^2$, which implies $z^2 - w^2 = 0$, so $(z - w)(z + w) = 0$.This means $z = w$ or $z = -w$.If $z = w$, then $|z| = |w|$, so $(z, w) \in A$.If $z = -w$, then $|z| = |-w| = |w|$, so $(z, w) \in A$.Thus, every element of $B$ is also an element of $A$, which means $B \subseteq A$.However, consider $(z, w) = (1, i)$. Here $|z| = |1| = 1$ and $|w| = |i| = 1$, so $|z| = |w|$, meaning $(1, i) \in A$.But $z^2 = 1^2 = 1$ and $w^2 = i^2 = -1$, so $z^2 
eq w^2$, meaning $(1, i) 
otin B$.Since $A$ contains elements not in $B$, $B \subset A$ is the correct relation.

Let $C$ denote the set of all complex numbers. Define $A = \{(z, w) \mid z, w \in C \text{ and } |z| = |w|\}$ and $B = \{(z, w) \mid z, w \in C \text{ and } z^2 = w^2\}$. Then:

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