Question

For a complex number z, \(\begin{array}{*{20}{c}} {lim}\\ {z \to i} \end{array}\frac{{{z^2} + 1}}{{{z^3} + 2z - i\left( {{z^2} + 2} \right)}}\) is
1. -2i
2. -i
3. i
4. 2i

Answer 1

Correct Answer - Option 4 : 2i

\(\begin{array}{*{20}{c}} {lim}\\ {z \to i} \end{array}\frac{{{z^2} + 1}}{{{z^3} + 2z - i\left( {{z^2} + 2} \right)}}\)

\(\begin{array}{*{20}{c}} {lim}\\ {z \to i} \end{array}\frac{{{z^2} - (i)^2}}{{z({z^2} + 2)- i\left( {{z^2} + 2} \right)}}\)

\(\begin{array}{*{20}{c}} {lim}\\ {z \to i} \end{array}\;\frac{{\left( {z + i} \right)\left( {z - i} \right)}}{{\left( {z - i} \right)\left( {{z^2} + 2} \right)}}\)

\( = \begin{array}{*{20}{c}} {lim}\\ {z \to i} \end{array}\frac{{\left( {z + i} \right)}}{{\left( {{z^2} + 2} \right)}} \)

\(= \frac{{i + i}}{{{i^2} + 2}} \)

= 2i