Question

In which quadrant of the complex plane, the point $\frac{(1+2i)}{(1-i)}$ lies?

• A. Fouth
• B. First
• C. Second
• D. Third

Answer 1

The correct option is C

Second

Determine the quadrant for the given point.

The point is: $\frac{(1+2i)}{(1-i)}$

Rationalizing the denominator,

$$\begin{gathered} \Rightarrow \frac{(1+2i)(1+i)}{(1-i)(1+i)} \\ \Rightarrow \frac{(1+3i+2i^2)}{1-i^2} \\ \Rightarrow \frac{(1-2+3i)}{2} \\ \Rightarrow \frac{(-1+3i)}{2} \end{gathered}$$

The point is now in the form of $x+iy$ where the $\text{'}x\text{'}$ part is negative and $\text{'}y\text{'}$ part is positive. Hence, the point $\frac{(1+2i)}{(1-i)}$ lies on the second quadrant, as in the second quadrant we have positive value of $\text{'}x\text{'}$, and negative values of $\text{'}y\text{'}$.

Therefore, the correct answer is Option (C).