Question

The real and imaginary part of the complex number (1-3i) / (1+2i) are:​

Answer 1

Answer:

Given ,

Complex number → ${\frac{(1-3i)}{(1+2i)}}$

$\large \bold{ \mathfrak{ Now,}}$

$\large \: = \frac{(1 - 3i)}{(1 + 2i)}$

$\large \: = \frac{(1 - 3i)(1 - 2i)}{(1 + 2i)(1 - 2i)}$

$\large \: = \frac{1 - 2i - 3i - 6}{1 + 4}$

$\large \: = \frac{ - 5 - 5i}{5}$

$\large \: = - 1 - i$

So,

• Real part → -1
• Imaginary part → -1

Answer 2

Answer:

Required real part is -1 and imaginary part is -1.

Step-by-step explanation:

Let a term be $\frac{x + iy}{a + ib}$

Now we want to remove imaginary part from the denominator part.

So,we are multiplying denominator and numerator by (a-ib).

$\frac{(x + iy)(a - ib)}{(a + ib)(a - ib)}$

$= \frac{xa - ibx + iay - {i}^{2}yb }{ {a}^{2} - {(ib)}^{2} } = \frac{ax - ibx + iay + yb}{ {a}^{2} + {b}^{2} } = \frac{ax + yb}{ {a}^{2} + {b}^{2} } + i \frac{ay - bx}{ {a}^{2} + {b}^{2} }$

By this process we can separate real part and imaginary part.

Given,$\frac{1 - 3i}{1 - 2i}$

We are multiplying denominator and numerator by (1+2i),

$\frac{(1 - 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)}$

$= \frac{1- 2i - 3i + 6 {i}^{2} }{ {1}^{2} - {2i}^{2} } = \frac{1 - 5i - 6}{(1 + 4)} = \frac{ - 5 - 5i}{5} = - \frac{5}{5} - i \frac{5}{5}$

So real part is -1 and imaginary part is -1.

Here applied formulas are

${i}^{2} = -1$

${x}^{2} - {y}^{2} = (x + y)(x - y)$.