If 5 + ix^3y^2 and x^3 + y^2 + 6i are conjuga | vedclass

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Question

$x^{3}+y^{2}=5 $ and $ x^{3} y^{2}=-6$

$\Rightarrow \mathrm{t}^{2}-5 \mathrm{t}-6=0$ has roots given by $\mathrm{x}^{3} $ and $ \mathrm{y}^{2}$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$x^{3}+y^{2}=5 $ and $ x^{3} y^{2}=-6$

$\Rightarrow \mathrm{t}^{2}-5 \mathrm{t}-6=0$ has roots given by $\mathrm{x}^{3} $ and $ \mathrm{y}^{2}$

or $t=6,-1$

$\Rightarrow x^{3}=-1 $ and $ y^{2}=6$

$ 	herefore  \arg (\mathrm{x}+\mathrm{iy})=	an ^{-1}(\pm \sqrt{6}) 	ext { or } 	an ^{2} 	heta=6$

List $I$	List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$	$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\| \ldots . .\left\|1-z_9\right\|}{10}$ equals	$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	$4.$ $2$

If $5 + ix^3y^2$ and $x^3 + y^2 + 6i$ are conjugate complex numbers and arg $(x + iy) = \theta $ , then ${\tan ^2}\,\theta $ is equal to

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