Question

pls help me to find the answer of this question​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \color{violet}{Re(z + z^{2} ) =Im(z) }}$

$\bold{ \leadsto \bf{ \pink{Let \: \: \: z = x + iy}}}$

So,

$\sf{ Re(x + iy + (x + iy)^{2} ) =Im(x + iy) }$

$\sf{ \implies Re(x + iy + (x) ^{2} +( iy)^{2} + 2 \cdot \: x \cdot \: iy ) =y }$

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

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

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

$\sf{ \implies x^{2} + x = {y}^{2} + y }$

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

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

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

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

$\sf{ \implies y = x \: \: \: \: \: or \: \: \: \: \: x + y + \dfrac{1}{2} + \dfrac{1}{2} = 0 }$

$\sf{ \implies y = x \: \: \: \: \: or \: \: \: \: \: x + y +1 = 0 }$