Complex Numbers Raised to a Power

Algebra Level 3

For which value of $N$ is the following true?

$(2+i)^{N}(3+i)^{N}+2500=0$

Clarification : $i =\sqrt{-1}$ .

1 solution

Tom Engelsman
Aug 30, 2016

Taking (2 + i)^N * (3 + i)^N = -2500, we can now write:

[(2+i)(3+i)]^N = -(2^2 * 5^4);

or (6 + 5i + i^2)^N = -(2^2 * 5^4);

or (5 + 5i)^N = -(2^2 * 5^4);

or 5^N * (1 + i)^N = -(2^2 * 5^4);

or 5^N * [sqrt(2) * exp(i * pi/4)]^N = -(2^2 * 5^4);

or 5^N * 2^(N/2) * exp(i * N*pi/4) = -(2^2 * 5^4)

which holds iff N = 4.