Question

if z=x+iy be such that that ampz-1/z-1=pi/4 then prove that x^2 +y^2 - 2y=1

Answer 1

x+iy)⅓ = a+ib

→ x+iy = (a+ib)³

→ x+iy = a³ + 3a²bi - 3ab² - ib³

→ (x + iy) = (a³-3ab²) + (3a² - b³)i

By equality of complex numbers:

x = a(a² - 3b²); y = b(3a² - b²)

[math]\dfrac{x}{a} = a^2 - 3b^2[/math] ….. (1)

[math]\dfrac{y}{b} = 3a^2 - b^2[/math] ….. (2)

Adding 1 and 2 we get

[math]\dfrac{x}{a} + \dfrac{y}{b} = 4(a^2 - b^2)[/math]

Proof of 1st part:

Let's think this logically. The conjugate of a comlex number is nothing but the reflection (mirror image) of the complex number about Re(Z) axis. So if one complex number is equal to the other complex number, then its reflection about the same axis should also be equal. So with that we can say that if two complex numbers are equal then their [math]([/math]conjugates are also equal. (Trying the theoretical proof. Will post soon as I get it)

Hope it helps.

Edit: Correct me if I am wrong

Edit 2: It will be [math](r_1)^\frac{1}{3} [/math]after coming out of the bracket and that will be equal to [math]r_2[/math]