Complex Number

Question

Answer 1

(i) Using the formula for the absolute value of a complex number we have

x + iy = |x + iy|

⇒ x + iy = \(\sqrt{x^2 + y^2}\)

⇒ x = \(\sqrt{x^2 + y^2}\) and y = 0.

Since y = 0 the equation x = \(\sqrt{x^2 + y^2}\) implies x = \(\sqrt{x^2}\) which implies x ≥ 0. Therefore, this equation is satisfied by

x ≥ 0, y = 0.

(ii) We compute 

\(\frac {x + iy}{x - iy} = x - iy\)

⇒ x + iy = (x - iy)²

⇒ x + iy = x² - y² - 2(xy)i.

This gives us two equations (since the real parts and imaginary parts must be equal),

x = x² - y²,

y = -2xy

If y ≠ 0 then from the second equation we have

\(x = - \frac 12\)

⇒ \(- \frac 12 = (- \frac 12)^2 - y^2\)

⇒ \(y = \frac {\sqrt 3}2\)

If y = 0 then we have x = x² so x = 0 or x = 1. But, x = 0 is not impossible since then \(\frac {x + iy}{x - iy} \) is undefined. Therefore we have two possibilities

x = 1, y = 0 or 

x = \(- \frac 12\),  y = \(\frac{\sqrt 3}2\)

(iii) Here we note that

Therefore, we have

Therefore, from the given equation we have

x + iy = 1

⇒ x = 1, y = 0.