Question

How do you write this in standard form? (6-7i)/(1-2i)

Answer 1

$\frac { \left( 6-7i \right) }{ \left( 1-2i \right) } \cdot 1\\ \\ =\frac { \left( 6-7i \right) }{ \left( 1-2i \right) } \cdot \frac { \left( 1+2i \right) }{ \left( 1+2i \right) } \\ \\ =\frac { 6+12i-7i-14{ i }^{ 2 } }{ 1+2i-2i-4{ i }^{ 2 } }$

$\\ \\ =\frac { 6+5i-14\left( -1 \right) }{ 1-4\left( -1 \right) } \\ \\ =\frac { 6+5i+14 }{ 1+4 } \\ \\ =\frac { 20+5i }{ 5 } \\ \\ =\frac { 20 }{ 5 } +\frac { 5i }{ 5 } \\ \\ =4+i$