+0

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

0
237
2

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

Guest Mar 8, 2015

#2
+18827
+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
+18827
+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

### 3 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details