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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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$

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

Similar Questions

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2+3 z+4 z^2}{2-3 z+4 z^2}$ is a real number, then the value of $|z|^2$ is. . . . .

Let $z$be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} \,(z) > 0$. Then $arg(z)$ is equal to

If $|z|\, = 4$ and $arg\,\,z = \frac{{5\pi }}{6},$then $z =$

The solution of the equation $|z| - z = 1 + 2i$ is

If $z$ is a complex number, then the minimum value of $|z| + |z - 1|$ is