Question

1- Resolva em C:

A) z2-2z+2=0
B) 2z^2+z+1=0
C) Z^2-iz+2=0
D) Z^4=1
E) z^4-81=0

Me ajudem por favor! É urgente.

Answer 1

a) $z^{2} -2z + 2 = 0$

Δ=b2−4ac
Δ=(2)2−4*(1)*(2)
Δ=4−8
Δ=−4

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(2) \pm \sqrt{-4}}{2*1} \\ \\ x = \frac{-2 \pm \sqrt{-4*i}}{2 } \\ \\ z' = \frac{-2 + 2*i}{2} \\ \\z' = \frac{-2- 2*i}{2} \\ \\ z' = -1 \\ \\ z'' = -1 \\ \\ S = \left \{- 1, - 1 \right.\}$

b) 
$2z^2 + z + 1 = 0$

Δ=b2−4ac
Δ=(1)2−4*(2)*(1)
Δ=1−8
Δ=−7

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-1 \pm \sqrt{-7}}{2*2} \\ \\ x = \frac{-1 \pm \sqrt{-7}*i}{4 } \\ \\ z' = \frac{-1 + 7*i}{4} \\ \\ z' =\frac{-1 + 7*i}{4} \\ \\ z'' =\frac{-1 + 7*i}{4} \\ \\ S = \left \{z' =\frac{-1 + 7*i}{4}, \frac{-1 - 7*i}{4} \right.\}$

c)

$z^{2} -i*z + 2 = 0$

(z - i) (z - 2i)
x = i
x = 2i
$S = \left \{i}, 2i \right.\}$

d)
$z^{4} = 1 \\ \\ z^{4} - 1 = 0 \\ \\ (z + 1)(z - 1)(z - i)(z - i) \\ \\ z' = -1 \\ \\ z'' = 1 \\ \\ z''' = -i \\ \\ z''''= i$

$S = \left \{1, - 1, -i, i} \right.\}$

e)

$z^{4} -81 = 0$

(z + 3)(z - 3)(z 3i)(z  - 3i)

z' = -3
z'' = 3
z''' = 3i
z'''' = -3i

$S = \left \{-3, 3, 3i, -3i} \right.\}$

Answer 2

E aí Erick,

lembre-se que √-1 = i:

$z^2-2z+2=0\\\\ \Delta=(-2)^2-4*1*2\\ \Delta=4-8\\ \Delta=-4\\\\ z= \dfrac{-(-2)\pm \sqrt{-4} }{2*1}= \dfrac{2\pm \sqrt{4}* \sqrt{-1} }{2}= \dfrac{2\pm2*i}{2}= \dfrac{2\pm2i}{2}=1\pm i\\\\\\ \boxed{S=\{1+i,~1-i\}}$

__________________

$2z^2+z+1=0\\\\ \Delta=1^2-4*2*1\\ \Delta=1-8\\ \Delta=-7\\\\ z= \dfrac{-1\pm \sqrt{-7} }{2*2}= \dfrac{-1\pm \sqrt{7}* \sqrt{-1} }{4}= \dfrac{-1\pm \sqrt{7}*i }{4}= \dfrac{-1\pm \sqrt{7}i }{4}\\\\\\ S=\left\{ -\dfrac{1}{4}+ \dfrac{ \sqrt{7} }{4}i,~ -\dfrac{1}{4}- \dfrac{ \sqrt{7} }{4}i\right\}$

__________________

$z^2-z+2=0\\\\ \Delta=(-1)^2-4*1*2\\ \Delta=1-8\\ \Delta=-7\\\\ z= \dfrac{-(-1)\pm \sqrt{-7} }{2*1}= \dfrac{1\pm \sqrt{7}* \sqrt{-1} }{2*1}= \dfrac{1\pm \sqrt{7} i}{2}\\\\\\ S=\left\{ \dfrac{1}{2}+ \dfrac{ \sqrt{7} }{2}i,~ \dfrac{1}{2}- \dfrac{ \sqrt{7} }{2}i\right\}$

__________________

$z^4=1\\ (z^2)^2=1\\\\ z^2=k\\\\ k^2=1\\ k=\pm \sqrt{1}\\ k=\pm1\\\\ para~k=1:~~~~~~~~~~~~~~~~para~k=-1\\ z^2=1~~~~~~~~~~~~~~~~~~~~~~~~z^2=-1\\ z=\pm \sqrt{1}~~~~~~~~~~~~~~~~~~~~~z=\pm \sqrt{-1} \\ z=\pm1~~~~~~~~~~~~~~~~~~~~~~~z=\pm \sqrt{1}* \sqrt{-1}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z= \pm1*i\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=\pm i\\\\\\ \boxed{S=\{1,-1,i,-i\}}$

___________________

$z^4-81=0\\ (z^2)^2-81=0\\\\ z^2=k\\\\ k^2-81=0\\ k^2=81\\ k= \pm\sqrt{81}\\ k=\pm9\\\\ z^2=k\\\\ z^2=9~~~~~~~~~~~~~~~~z^2=-9\\ z=\pm \sqrt{9}~~~~~~~~~~~~~z=\pm \sqrt{-9} \\ z=\pm3~~~~~~~~~~~~~~~z=\pm3i\\\\\\ \boxed{S=\{3,-3,3i,-3i\}}$

Tenha ótimos estudos =))