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

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

1. 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

   Reply

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

   Let:

   z' = 1 - 4i (conjugate)

   So:

   [tex]\frac{3 - 2i}{1 + 4i} . \frac{1 - 4i}{1 - 4i}[/tex]

   [tex]\frac{3 - 12i - 2i + 8i^2}{1 - 4i + 4i - 16i^2}[/tex]

   [tex]\frac{3 - 14i + 8( \sqrt{-1})^2 }{1 - 16( \sqrt{-1})^2 }[/tex]

   [tex]\frac{3 - 14i + 8(-1)}{1 - 16(-1)}[/tex]

   [tex]\frac{3 - 14i - 8}{1 + 16}[/tex]

   [tex]\frac{- 5 - 14i}{17}[/tex]

   It's done, all steps have been made.

   Hope it helps.

   Reply

3. [tex]\frac{3-2i}{i+4}[/tex] is written in the form of a+bi as [tex]\frac{-11i}{17}+\frac{10}{17}[/tex]

   Step-by-step explanation:

   As given the expression in the question be

   [tex]= \frac{3-2i}{i+4}[/tex]

   Multiply denominator and numerator by i-4 .

   [tex]= \frac{3-2i\times i-4}{i+4\times i-4}[/tex]

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

   Apply this in the above

   [tex]= \frac{3-2i\times i-4}{i^{2} - 4^{2}}[/tex]

   [tex]= \frac{3i-12-2i^{2}+8i}{i^{i}- 16}[/tex]

   (As i² = -1 )

   [tex]= \frac{11i-12-2\times -1}{-1 - 16}[/tex]

   [tex]= \frac{11i-12+2}{-1 - 16}[/tex]

   [tex]= \frac{11i-10}{-17}[/tex]

   [tex]= \frac{-11i}{17}+\frac{10}{17}[/tex]

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

   Reply

4. 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. =)

   Reply

5. 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.

   [tex]\frac{3-2i}{1+4i}= \frac{(3-2i)}{(1+4i)} \frac{(1-4i)}{(1-4i)}= \frac{(3-2i)(1-4i)}{(1+4i)(1-4i)}[/tex]

   [tex]\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[/tex]

   Reply

6. [tex]\dfrac{x+iy}{u+iv}\times\dfrac{u-iv}{u-iv}=\dfrac{(x+iy)(u-iv)}{u^2-(iv)^2}[/tex]
   [tex]=\dfrac{xu-i^2yv+iyu-ixv}{u^2+v^2}[/tex]
   [tex]=\underbrace{\dfrac{xu+yv}{u^2+v^2}}_a+i\underbrace{\dfrac{yu-xv}{u^2+v^2}}_b[/tex]

   Reply