Question

please help me. I need urgently that's answer​

Answer 1

Step-by-step explanation:

5.

We have,

$\tt{ \bar{z} = i \: z^{2} }$

Let $\sf{\pink{z=x+i\,y}}$

So,

$\sf{ x - i y = i \: (x + iy)^{2} }$

$\sf{ \implies x - i y = i \: \left \{(x)^{2} +( iy)^{2} + 2xyi \right \} }$

$\sf{ \implies x - i y = i \: \left \{x^{2} - y^{2} + 2xyi \right \} }$

$\sf{ \implies x - i y = i \: (x^{2} - y^{2}) + 2xy \: i^{2} }$

$\sf{ \implies x - i y = - 2xy + i \: (x^{2} - y^{2}) }$

On compairing,

$\sf{x = - 2xy \: \: \: \: \: \: and \: \: \: \: \: \: - y = {x}^{2} - {y}^{2} }$

$\sf{ \implies \: 1= - 2y \: \: \: \: \: \: and \: \: \: \: \: \: y^{2} = {x}^{2} + y }$

$\sf{ \implies \: y= - \dfrac{1}{ 2} \: \: \: \: \: \: and \: \: \: \: \: \: y^{2} = {x}^{2} + y }$

Put the value of y in 2nd equation,

$\sf{ \implies \left( - \dfrac{1}{2}\right)^{2} = {x}^{2} + \left( - \dfrac{1}{2}\right) }$

$\sf{ \implies \dfrac{1}{4} + \dfrac{1}{2} = {x}^{2} }$

$\sf{ \implies {x}^{2} = \dfrac{3}{4}}$

$\sf{ \implies x = \pm\dfrac{ \sqrt{3}}{2}}$

Hence, required complex numbers are

$\sf{z = \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2}i \: \: \: \: \: \: and \: \: \: \: \: \: z = - \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2}i }$