# Multiply and express the answer in a+bi form : (8-4i)(6+i)

Multiply and express the answer in a+bi form : (8-4i)(6+i)

## This Post Has 6 Comments

1. sammamamam1090 says:

I hope this helps you

1^2+2.1.3i+(3i)^2

1+6i+9i^2

1+6i+9. (-1)

1+6i-9

-8+6i

2. ayoismeisjjjjuan says:

When dividing complex numbers, we multiplicate both numerator and denominator by the conjugate of the denominator, like this:

Let:

z' = 1 - 4i (conjugate)

So:

$\frac{3 - 2i}{1 + 4i} . \frac{1 - 4i}{1 - 4i}$

$\frac{3 - 12i - 2i + 8i^2}{1 - 4i + 4i - 16i^2}$

$\frac{3 - 14i + 8( \sqrt{-1})^2 }{1 - 16( \sqrt{-1})^2 }$

$\frac{3 - 14i + 8(-1)}{1 - 16(-1)}$

$\frac{3 - 14i - 8}{1 + 16}$

$\frac{- 5 - 14i}{17}$

It's done, all steps have been made.

Hope it helps.

3. fredo99 says:

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

4. joneskc0629 says:

Multiply the complex numbers:

(8 – 4i) · (6 + i)

= (8 – 4i) · 6 + (8 – 4i) · i

= 8 · 6 – 4i · 6 + 8 · i – 4i · i

= 48 – 24i + 8i – 4i²          (but, i² = – 1)

= 48 – 24i + 8i – 4 · (– 1)

= 48 – 24i + 8i + 4

= 48 + 4 – 24i + 8i

= 52 – 16i    <————    this is the answer.

I hope this helps. =)

5. eggg65 says:

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.

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

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

$\dfrac{x+iy}{u+iv}\times\dfrac{u-iv}{u-iv}=\dfrac{(x+iy)(u-iv)}{u^2-(iv)^2}$
$=\dfrac{xu-i^2yv+iyu-ixv}{u^2+v^2}$
$=\underbrace{\dfrac{xu+yv}{u^2+v^2}}_a+i\underbrace{\dfrac{yu-xv}{u^2+v^2}}_b$