Math, asked by anu1545, 3 months ago

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

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Answered by senboni123456
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 }

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