y4+1=0 has only complex solutions.
y4=-1, y2=±i⇒(y2-i)(y2+i)=0=y4+1.
Let y=a+ib where a and b are real constants.
y2=a2-b2+2aib=±i, y=±√i.
We have two equations to solve for a and b:
(1) a2-b2+2aib=i; therefore a2-b2=0⇒a=±b but 2ab=1 and a and b must both be real, when we equate the real and imaginary parts. So:
a=b and 2a2=2b2=1, making a=b=1/√2, also written √2/2. So y=√2/2+i√2/2. Also a=b=-√2/2, making y=-√2/2(1+i).
(2) a2-b2+2aib=-i; therefore a=-b and -2a2=-1, a=√2/2, b=-√2/2, y=√2/2-i√2/2. Also y=-√2/2(1-i), because a=-√2/2, b=√2/2.
The four roots are:
±√2/2(1+i), ±√2/2(1-i)⇒
(y+√2/2(1+i))(y-√2/2(1+i))(y+√2/2(1-i))(y-√2/2(1-i))=y4+1=0.
(y2-½(1+2i-1))(y2-½(1-2i-1)=(y2-i)(y2+i)=y4+1=0.