Test Index

AP EAPCET 06-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 8 of 160

Marks: +1, -0

z

\overline{z}=i z^{2}

Solution: 👈: Video Solution

Let

\dot{z}=x+iy

\overline{z}\;=x-iy \Rightarrow \overline{z}=i z^{2}

\Rightarrow x-iy\;=i[(x+iy)(x+iy)]

\Rightarrow x-iy\;=i[x^{2}+2ixy+i^{2}y^{2}]

\Rightarrow x-iy\;=i[x^{2}-y^{2}+2ixy]

\Rightarrow x-iy\;=-2xy+i(x^{2}-y^{2})

Comparing the real and imaginary part, we get

-2xy=x

\Rightarrow\; \text{Either}\; x\;=0\; \text{or}\; y=-\;\frac{1}{2}

x^{2}-y^{2}\;=-y \Rightarrow x^{2}=y^{2}-y

x=0

, then

y=0

y=1

y=-\;\frac{1}{2}

, then

x^{2}=\;\frac{1}{4}+\;\frac{1}{2}=\;\frac{3}{4}

x=\pm \;\frac{\sqrt{3}}{4}

There are four solutions, i.e.

(0,0),(0,1), \left(\;\frac{\sqrt{3}}{2},-\;\frac{1}{2}\right)

and

\left(-\;\frac{\sqrt{3}}{2},-\;\frac{1}{2}\right)

Go to Question:

Prev Question

Next Question