Help with imaginary/complex numbers

[Eagle43]

Hello, could anyone help with this problem: [imath]z^4=-4[/imath]? I got to *work shown below*, before realizing that I would just get the same equation I started with

 

[Deleted member 4993]

Hello, could anyone help with this problem: [imath]z^4=-4[/imath]? I got to *work shown below*, before realizing that I would just get the same equation I started with

View attachment 28705

What are you supposed to do with:

z⁴ = -4

What is the COMPLETE question - what is the objective?

 

[Dr.Peterson]

Assuming the goal is to solve for z, it seems that your next step is to solve [imath]z^2=2i[/imath] for z. You don't do that by squaring! You need to take the square root.

We don't know what you have learned about finding such roots. Depending on what you know, you might just write [imath]z=x+iy[/imath] and solve for x and y; or use polar form and/or De Moivre's formula.

 

[Eagle43]

Ok...so are the answers to that [imath]z=\sqrt 2i, z=-\sqrt2i[/imath]? And then I would just simplify? Or in other words, I could take the fourth root of [imath]-4[/imath]?
What about the other situation - [imath]z^2+2i=0[/imath] could I make that to [imath]z^2=-2i[/imath]?
Also, sorry for not specifying, @Subhotosh Khan, the objective is to "Find all complex numbers

such that

"

 

[pka]

Hello, could anyone help with this problem: [imath]z^4=-4[/imath]? I got to *work shown below*, before realizing that I would just get the same equation I started with

The number [imath]-4[/imath] has four complex fourth roots.
In polar form [imath]-4=4\exp\left(\pi i\right)[/imath] so that the principal fourth root is [imath]\rho=\sqrt{2}\exp\left(\dfrac{\pi i}{4}\right) [/imath].
Now if [imath]\theta=\exp\left(\dfrac{\pi i}{2}\right)[/imath] then the four fourth of [imath]-4[/imath] are [imath]\rho\cdot\theta^k,~k=0,1,2,3[/imath] SEE HERE

 

[Eagle43]

Ok. I have one more question - if I know that [imath](z^2)^2=(-2i)^2[/imath] can I square root the whole equation without changing the value?

 

[Dr.Peterson]

Ok. I have one more question - if I know that [imath](z^2)^2=(-2i)^2[/imath] can I square root the whole equation without changing the value?

Presumably you mean, take the square root of both sides without changing the solution set. Words matter.

The answer is, you can, if you take account of both possible square roots! Just dropping the exponent is not enough.

But there are better ways, depending on what you have learned (which we are still not sure of).

Ok...so are the answers to that [imath]z=\sqrt 2i, z=-\sqrt2i[/imath]? And then I would just simplify? Or in other words, I could take the fourth root of [imath]-4[/imath]?
What about the other situation - [imath]z^2+2i=0[/imath] could I make that to [imath]z^2=-2i[/imath]?

No, you can't just take the square root of 2. It would have to be [imath]z=\sqrt{2i}, z=-\sqrt{2i}[/imath]. Do you see the difference? And do you know the square root(s) of i?