Question

plz answer me fast ​

Answer 1

Answer:

$\dfrac{\left(1+2i\right)\left(1+3i\right)}{2+i}=-1+3i$

Step-by-step explanation:

$\dfrac{\left(1+2i\right)\left(1+3i\right)}{2+i}$

$\black{\left(1+2i\right)\left(1+3i\right):}$

$\left(1+2i\right)\left(1+3i\right)$

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

$\gray{a=1,\:b=2,\:c=1,\:d=3}$

$=\left(1\cdot \:1-2\cdot \:3\right)+\left(1\cdot \:3+2\cdot \:1\right)i$

$\gray{\mathrm{Refine}}$

$=-5+5i$

$=\dfrac{-5+5i}{2+i}$

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

$\gray{a=-5,\:b=5,\:c=2,\:d=1}$

$=\dfrac{\left(-5\cdot \:2+5\cdot \:1\right)+\left(5\cdot \:2-\left(-5\right)\cdot \:1\right)i}{2^2+1^2}$

$\gray{\mathrm{Refine}}$

$=\dfrac{-5+15i}{5}$

$\black{\mathrm{Simplify}\:\dfrac{-5+15i}{5}:}$

$\dfrac{-5+15i}{5}$

$\gray{\mathrm{Factor}\:-5+15i:\quad 5\left(-1+3i\right)}$

$=\dfrac{5\left(-1+3i\right)}{5}$

$\gray{\mathrm{Divide\:the\:numbers:}\:\frac{5}{5}=1}$

$=-1+3i$

Answer 2

Answer:

Hey mate please refer to the attachment

here we used (a+b)(a-b)=a^2-b^2

and we also know that i^2= -1