Question

1+2i+3i2÷1-2i+3i2
Solve by a+ib

Answer 1

Answer:

$\dfrac{1+2i+3i^2}{1-2i+3i^2}=-i$

Step-by-step explanation:

$\dfrac{1+2i+3i^2}{1-2i+3i^2}$

$\black{1+2i+3i^2}$

$\gray{3i^2=-3}$

$=1+2i-3$

$\gray{\mathrm{Subtract\:the\:numbers:}\:1-3=-2}$

$=-2+2i$

$=\dfrac{-2+2i}{1-2i+3i^2}$

$\black{1-2i+3i^2}$

$\gray{3i^2=-3}$

$=1-2i-3$

$\gray{\mathrm{Subtract\:the\:numbers:}\:1-3=-2}$

$=-2-2i$

$=\dfrac{-2+2i}{-2-2i}$

$\black{\mathrm{Factor}\:-2+2i:\quad 2\left(-1+i\right)}$

$=\dfrac{2\left(-1+i\right)}{-2-2i}$

$\black{\mathrm{Factor}\:-2-2i:\quad -2\left(1+i\right)}$

$=-\dfrac{2\left(-1+i\right)}{2\left(1+i\right)}$

$\gray{\mathrm{Divide\:the\:numbers:}\:\dfrac{2}{2}=1}$

$=-\dfrac{-1+i}{\left(1+i\right)}$

$\gray{\mathrm{Apply\:complex\:arithmetic\:rule}:\quad \dfrac{a+bi}{c+di}\:=\:\dfrac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)}\:=\:\dfrac{\left(ac+bd\right)+\left(bc-ad\right)i}{c^2+d^2}}$

$\gray{a=-1,\:b=1,\:c=1,\:d=1}$

$=-\dfrac{\left(-1\cdot \:1+1\cdot \:1\right)+\left(1\cdot \:1-\left(-1\right)\cdot \:1\right)i}{1^2+1^2}$

$\gray{\mathrm{Refine}}$

$=-\dfrac{2i}{2}$

$\gray{\mathrm{Simplify}\:-\dfrac{2i}{2}}$

$=-i$