Question

Assunto: Forma algébrica de um número complexo.​

Answer 1

1)

$z_{1}=1+2i\\z_{2}=-1+3i\\z_{3}=3-2i$

a)

$\bf{z_{1}+z_{2}=\cancel1+2i-\cancel1+3i=5i}$

b)

$\bf{z_{1}-z_{2}=(1+2i)-(-1+3i)}$

$\bf{z_{1}-z_{2}=1+2i+1-3i}$

$\bf{z_{1}-z_{2}=2-i}$

c)

$z_{1}.z_{2}=(1+2i)(-1+3i) =-1+3i-2i+6{i}^{2}$

$z_{1}.z_{2}=-1+i+6(-1)$

$\bf{z_{1}.z_{2}=i-7}$

d)

$(z_{1}+z_{2}).z_{3}=5i(3-2i) =15i-6{i}^{2}$

$(z_{1}+z_{2}).z_{3}=15i-6(-1)$

$\bf{(z_{1}+z_{2}).z_{3}=15i+6}$

2)

a) faça a parte real igual a zero

${x}^{2}-x=0\\x(x-1)=0$

$\bf(x=0}\\\bf{x-1=0}\\\bf{x=1}$

b) faça a parte imaginária igual a zero

$3+({x}^{2}-4)i$

${x}^{2}-4=0\\{x}^{2}=4\\x=±\sqrt{4}$

$\bf{x=-2}\:e\:x=2$

3)

a)

$4{x}^{2}-4x+2=0$

$\Delta=16-32=-16$

$x=\dfrac{4±4i}{8}$

$x'=\dfrac{4+4i}{8}=\dfrac{\cancel{4}(1+i)}{\cancel{8}}$

$\bf{x'=\dfrac{1+i}{2}}$

$\bf{x''=\dfrac{1-i}{2}}$

b)

${x}^{2}-14x+50=0$

$\Delta=196-200=-4$

$x=\dfrac{14±2i}{2}=\dfrac{\cancel2(7±i)}{\cancel2}$

$\bf{x'=7+i}\\\bf{x''=7-i}$