METHOD 1
4th root is found by taking the square root twice. Because -81 is a negative number its square root is complex=9i or more correctly ±9i.
If we take the square root again we need to find √-9i and √9i.
If √-9i=a+ib where a and b are real,
-9i=a2-b2+2aib.
So a2=b2, a=±b; 2ab=-9, so a=-b and 2a2=9, a=±3/√2=±3√2/2 and b=∓3√2/2.
√-9i=3√2/2-3i√2/2 or -3√2/2+3i√2/2, which can be written ±3√2/2(1-i).
If √9i=p+iq, 9i=p2-q2+2piq; p2=q2, 2pq=9⇒p=q=±3√2/2.
√9i=±3√2/2(1+i).
CHECK
[±3√2/2(1-i)]2=(9/2)(1-1-2i)=-9i. So √-9i=±3√2/2(1-i).
[±3√2/2(1+i)]2=(9/2)(1-1+2i)=9i. So √9i=±3√2/2(1+i).
METHOD 2
Let z=aeiθ=a(cosθ+isinθ) and z4=-81=a4e4iθ=a4(cos(4θ)+isin(4θ)).
a=±3, so cos(4θ)+isin(4θ)=-1. sin(4θ)=0 and cos(4θ)=-1⇒4θ=(2n+1)π, θ=¼(2n+1)π.
0≤θ≤2π is the range for θ⇒0≤n≤3⇒θ=π/4, 3π/4, 5π/4, 7π/4⇒sinθ=±√2/2.
z=3√2/2(1+i), -3√2/2(1+i), 3√2/2(1-i), -3√2/2(1-i).