Complex Numbers - Number of Solutions

[Lancelot1]

Hiya all,

I need your assistance with the following problem:

A) Show that the equation

\[z^{2}+i\bar{z}=(-2)\]

has only two imaginary solutions.

B) If Z1 and Z2 are the solutions, draw a rectangle which has the following vertices:

Z1+3 , Z2+3 , Z1+i , Z2+i

I do not know how to even start. Should I try to write Z as a+bi ? Please help (Doh)

 

[Opalg]

A) Show that the equation
\[z^{2}+i\bar{z}=(-2)\]
has only two imaginary solutions.

B) If Z1 and Z2 are the solutions, draw a rectangle which has the following vertices:

Z1+3 , Z2+3 , Z1+i , Z2+i

I do not know how to even start. Should I try to write Z as a+bi ?

Yes, write $z = a+ib$. Then the equation becomes $(a+ib)^2 + i(a-ib) + 2 = 0.$ Now remember that if a complex number is zero then its real and imaginary parts must both be zero. That will give you two equations for the real numbers $a$ and $b$, and you should find that there are just two solutions.

I don't know what to say about part B), because as far as I can see, those four points do not form the vertices of a rectangle. (I think it should be a parallelogram.)

 

[Lancelot1]

Isn't a rectangle a sort of parallelogram ? How did you see it's a parallelogram ?

 

[HallsofIvy]

Yes, a rectangle is a type of parallelogram. Because you said it is a rectangle, it is a parallelogram. I'm not sure that calling it a parallelogram helps though!