# Help with the next complex variable limit?

## ( (Imz^2 )/"z+Re(z)" where $z \to 1 - i$

Jul 12, 2018

$= \frac{- 4 - 2 i}{5}$

#### Explanation:

$f \left(z\right) = \frac{\mathbb{I m} \left({z}^{2}\right)}{z + \mathbb{R e} \left(z\right)}$

$\therefore f \left(1 - i\right) = \frac{- 2}{\left(2 - i\right)}$

$= \frac{- 2 \left(2 + i\right)}{\left(2 - i\right) \left(2 + i\right)} = \frac{- 4 - 2 i}{5}$

The function is:

$f \left(z\right) = \frac{2 x y}{2 x + i y}$

$= \frac{4 {x}^{2} y}{4 {x}^{2} + {y}^{2}} - i \frac{2 x {y}^{2}}{4 {x}^{2} + {y}^{2}}$

By inspection, the function is continuous at all points [except maybe the origin]. So the limit is simply the value of the function at the point in question, ie $f \left(1 - i\right)$.