What is the solution of $(m^{2}+1)^2=0?$

Question

I understand the solution of $m^{2}+1=0$ is $\iota$. However for sure this solution ($(m^{2}+1)^2=0$) should contain four roots. The answer reads $\pm \iota$ and $\pm \iota$. However I am not able to calculate all the roots. I can arrive at most up to $\pm \iota$ and cannot proceed any further.

Answer 1

Here the factorisation which, I hope, will make you understand why there are really $4$ roots: $$(m^2+1)^2=\bigr((m-i)(m+i)\bigl)^2=\underbrace{(m-i)}_{\text{two roots}}{}^2\underbrace{(m+i)}_{\text{two roots}}{}^2.$$ You have $4$ factors, which are equal in pairs.

Answer 2

Why should it have four roots ? A $4-$degree polynomial has at most four distinct roots...

Here $(m^2+1)^2=0$ implies that $m^2+1=0$, then you have $m=i$ or $m=-i$ in $\mathbb{C}$.

Answer 3

Hint: $(m^2+1)^2=(m^2+1)(m^2+1)=(m+i)(m-i)(m+i)(m-i)$

Answer 4

The solution is obtained by solving $m^{2} + 1 = 0$ which yields $m = \pm i$.

That being said, you have effectively a quartic equation which will have exactly four roots counting multiplicities (repititions).

Here, $-i$ and $i$ are the two distinct solutions, each having multiplicity two.