Solving complex equation

[steve.b]

1)
z[sup:3siwheic]2[/sup:3siwheic]+2iz+1+i=0

2)
z[sup:3siwheic]6[/sup:3siwheic]-z[sup:3siwheic]3[/sup:3siwheic]+1-i=0

Thank you!

 

[Deleted member 4993]

1)
z[sup:1cdizupv]2[/sup:1cdizupv]+2iz+1+i=0

2)
z[sup:1cdizupv]6[/sup:1cdizupv]-z[sup:1cdizupv]3[/sup:1cdizupv]+1-i=0

Thank you!

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you

 

[steve.b]

Ok, unfortunatelly, I am not quite sure, but I started and I don't really know, if it is correct:

1)
z[sup:2ob6jfch]2[/sup:2ob6jfch]+2iz+1+i=0
z[sub:2ob6jfch]1,2[/sub:2ob6jfch]=-[2i+[(2i)[sup:2ob6jfch]2[/sup:2ob6jfch]-4-4i][sup:2ob6jfch]0.5[/sup:2ob6jfch]]/2,
so (a+bi)[sup:2ob6jfch]2[/sup:2ob6jfch]=-4-4i
a[sup:2ob6jfch]2[/sup:2ob6jfch]-b[sup:2ob6jfch]2[/sup:2ob6jfch]+2abi=-4-4i
a[sup:2ob6jfch]2[/sup:2ob6jfch]-b[sup:2ob6jfch]2[/sup:2ob6jfch]=-4-4i
2ab=-b
==> 4/b[sup:2ob6jfch]2[/sup:2ob6jfch]-b[sup:2ob6jfch]2[/sup:2ob6jfch]=-4
b[sup:2ob6jfch]2[/sup:2ob6jfch][sub:2ob6jfch]1,2[/sub:2ob6jfch]=4.83 or -0.83 (it can't) so b[sub:2ob6jfch]1[/sub:2ob6jfch]=2.2==> a[sub:2ob6jfch]1[/sub:2ob6jfch]=-1.29 b[sub:2ob6jfch]2[/sub:2ob6jfch]=-2,2==> a[sub:2ob6jfch]2[/sub:2ob6jfch]=1,29
so z[sub:2ob6jfch]1[/sub:2ob6jfch]=[-2i-1.29+2.2i]/2=-0,64+0,1i z[sub:2ob6jfch]2[/sub:2ob6jfch]=+0,64+1,6i
But I don't think that it is the good solution, so please help me, where is the mistake!

 

[Deleted member 4993]

1)
z[sup:5bngf27m]2[/sup:5bngf27m]+2iz+1+i=0

\(\displaystyle z_{1,2} \ = \ \frac{-2i \pm\sqrt{(2i)^2 - 4(1+i)}}{2}\)

\(\displaystyle z_{1,2} \ = \ i \pm i \sqrt{2+i}\)

now continue....

2)
z[sup:5bngf27m]6[/sup:5bngf27m]-z[sup:5bngf27m]3[/sup:5bngf27m]+1-i=0

For this one slove for z[sup:5bngf27m]3[/sup:5bngf27m] first

Thank you!

 

[steve.b]

Thank you so much!