Home Mathematics 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)Mathematics Live4dramaoy0yf9October 23, 20216 CommentsMultiply and express the answer in a+bi form : (8-4i)(6+i)
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
[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
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
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
[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
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
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.
[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 .
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. =)
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]
[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]