Question

Simplify the complex number. Express your answer in a + bi form and include each step necessary in simplifying. 3-2i/1+4i

Answer 1

Answer:

$\frac{3-2i}{i+4}$ is written in the form of a+bi as $\frac{-11i}{17}+\frac{10}{17}$

Step-by-step explanation:

As given the expression in the question be

$= \frac{3-2i}{i+4}$

Multiply denominator and numerator by i-4 .

$= \frac{3-2i\times i-4}{i+4\times i-4}$

(As by using the property  (a-b)(a+b)= (a² - b²))

Apply this in the above

$= \frac{3-2i\times i-4}{i^{2} - 4^{2}}$

$= \frac{3i-12-2i^{2}+8i}{i^{i}- 16}$

(As i² = -1 )

$= \frac{11i-12-2\times -1}{-1 - 16}$

$= \frac{11i-12+2}{-1 - 16}$

$= \frac{11i-10}{-17}$

$= \frac{-11i}{17}+\frac{10}{17}$

This is the representation of $\frac{3-2i}{i+4}$ is written in the form of a+bi .

Answer 2

1. 
When we divide two complex numbers, we multiply by the conjugate of the complex number in the denominator.

2. Remark: The conjugate of a+bi is a-bi.

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

$\frac{(3-2i)(1-4i)}{(1+4i)(1-4i)}= \frac{3-4i-2i+(2i)(4i)}{1-(4i)^{2} } = \frac{3-6i-8}{1+16}= \frac{-5-6i}{17}= \frac{-5}{17}- \frac{6}{17}i$