What is the fourth root of -81?

Question

( -81) raised to the (1/4) power

Answer 1

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=ae^iθ=a(cosθ+isinθ) and z⁴=-81=a⁴e^4iθ=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).