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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 8 of 160

Marks: +1, -0

z

(7+i)(z+\overline{z})-(4+i)(z-\overline{z})+116i=0

z \overline{z} =

Solution:

Consider a complex equation,

z=x+iy

It conjugate is

\overline{z}=x-iy

\text{Now}

(7+i)(z+\overline{z})-(4+i)(z-\overline{z})+116i = 0

(7+i)(2x)-(8iy-2y)+116i = 0

(14x+2y)+i(2x-8y+116) = 0

The equations obtained are

2x-8y+116 = 0

14x+2y = 0

From above two equations,

x=-2, y=14

The modulus is,

\left|z\right| = \sqrt{x^{2}+y^{2}}

= \sqrt{(-2)^{2}+14^{2}}

= \sqrt{200}

\text{Thus, } z \overline{z} = \left|z\right|^{2} = (\sqrt{200})^{2} = 200

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