+0  
 
0
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$$\left({\frac{\left({\mathtt{4}}\left[{{i}}^{{\mathtt{6}}}\right]{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{i}\right)}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{2}}{i}\right)}}\right){\mathtt{\,-\,}}\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{i}\right)$$=???

Guest Mar 8, 2015

Best Answer 

 #2
avatar+20011 
+5

$$\small{\text{
$
\boxed{
\dfrac{4i^6+3i } {1-2i}
-(9+4i) =\ ?
}
$
}}\\\\
\small{\text{
$
i^2 = -1
\qquad i^4 = i^2 \cdot i^2 = (-1)\cdot (-1) = 1
\qquad i^6 = i^2 \cdot i^2 \cdot i^2 = (-1)\cdot (-1) \cdot (-1) = -1
$
}}\\$$

we set

$$\small{\text{$i^6 = (-1)^3 = -1$}}$$

so we have:

$$\small{\text{
$
\dfrac{4i^6+3i } {1-2i}
-(9+4i) \quad | \quad \boxed{i^6=-1}
$
}}\\\\
\small{\text{
$
=\dfrac{-4+3i } {1-2i}
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { (1-2i)\codt(1+2i) }
-(9+4i) \quad | \quad \boxed{(a-ib)\cdot (a+ib) = a^2+b^2}
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 1^2+2^2 }
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 5 }
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 5 }
-5\cdot\dfrac{ (9+4i) }{5}
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i)-5\codt(9+4i) } { 5 }
$
}}\\\\
\small{\text{
$
=\dfrac{ -4-8i+3i+6i^2 -45 -20i } { 5 }
\quad | \quad \boxed{i^2=-1}
$
}}\\\\
\small{\text{
$
=\dfrac{ -4 -8i + 3i +6\cdot (-1) -45 -20i } { 5 }
$
}}\\\\
\small{\text{
$
=\dfrac{ -4 -8i + 3i -6 -45 -20i } { 5 }
\quad $ we put together
}}\\\\
\small{\text{
$
=\dfrac{ -55 -25i } { 5 }
$
}}\\\\
\small{\text{
$
=-11 -5i
$
}}$$

heureka  Mar 8, 2015
 #1
avatar
+5

Remember that i2= -1

i6=(i2)3 ; do this first ,combine like terms and then multply by the conjugate of the denominator, (1+2i).

Should be simple after that, keep combining like terms. I got -11-3i.

Guest Mar 8, 2015
 #2
avatar+20011 
+5
Best Answer

$$\small{\text{
$
\boxed{
\dfrac{4i^6+3i } {1-2i}
-(9+4i) =\ ?
}
$
}}\\\\
\small{\text{
$
i^2 = -1
\qquad i^4 = i^2 \cdot i^2 = (-1)\cdot (-1) = 1
\qquad i^6 = i^2 \cdot i^2 \cdot i^2 = (-1)\cdot (-1) \cdot (-1) = -1
$
}}\\$$

we set

$$\small{\text{$i^6 = (-1)^3 = -1$}}$$

so we have:

$$\small{\text{
$
\dfrac{4i^6+3i } {1-2i}
-(9+4i) \quad | \quad \boxed{i^6=-1}
$
}}\\\\
\small{\text{
$
=\dfrac{-4+3i } {1-2i}
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { (1-2i)\codt(1+2i) }
-(9+4i) \quad | \quad \boxed{(a-ib)\cdot (a+ib) = a^2+b^2}
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 1^2+2^2 }
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 5 }
-(9+4i)
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i) } { 5 }
-5\cdot\dfrac{ (9+4i) }{5}
$
}}\\\\
\small{\text{
$
=\dfrac{ (-4+3i) \codt(1+2i)-5\codt(9+4i) } { 5 }
$
}}\\\\
\small{\text{
$
=\dfrac{ -4-8i+3i+6i^2 -45 -20i } { 5 }
\quad | \quad \boxed{i^2=-1}
$
}}\\\\
\small{\text{
$
=\dfrac{ -4 -8i + 3i +6\cdot (-1) -45 -20i } { 5 }
$
}}\\\\
\small{\text{
$
=\dfrac{ -4 -8i + 3i -6 -45 -20i } { 5 }
\quad $ we put together
}}\\\\
\small{\text{
$
=\dfrac{ -55 -25i } { 5 }
$
}}\\\\
\small{\text{
$
=-11 -5i
$
}}$$

heureka  Mar 8, 2015

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