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# 4i^6+3i/1-2i-(9+4i)

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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

#2
+19207
+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
Sort:

#1
+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
+19207
+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

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