First slide

Algebra of complex numbers

Question

Suppose $z_{1}, z_{2}$ are two distinct complex numbers and $a, b$ are real numbers.
Statement $- 1 :$ If $z_{1} + z_{2} = a, z_{1} z_{2} = b,$ then $a r g (z_{1} z_{2}) = 0$
Statement $- 2 :$ If $z_{1} + z_{2} = a, z_{1} z_{2} = b,$ then $\bar{z_{1}} = z_{2}$

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Statement- $2$ is false. This can be seen by taking $z_{1}$ and $z_{2}$ to be distinct real numbers.
For statement $- 1$ , take $z_{1 =} α_{1} + i β_{1}, z_{2} = α_{2} + i β_{2} .$ We have
$β_{1} + β_{2} = 0, α_{1} β_{2} + α_{2} β_{1} = 0$
If $β_{2} = 0,$ then $β_{1} = 0,$ and statement $- 1$ is true.
If $β_{2} \neq 0,$ then $β_{1} = - β_{2}$ and $(α_{1} - α_{2}) β_{2} = 0$
$\Rightarrow$ $α_{1} - α_{2} = 0$ or $α_{1} = α_{2}$
Thus, $α_{2} + i β_{2} = α_{1} - i β_{1} \Rightarrow z_{2} = \bar{z_{1}}$
In this case also $\begin{array}{r} \arg (z_{1} z_{2}) = \arg (z_{1} {\bar{z}}_{1}) \\ = \arg ({|z_{1}|}^{2}) = 0 \end{array}$

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