Asked by maths-lover
solve for x? if x^2+j16=0 where j=sqrt(-1)
Answers
Answered by
Steve
unusual to have complex coefficients, but
x^2 = -16j = 16cis(3π/2)
x = ±4 cis(3π/4)
= ±4 (-1/√2 + 1/√2 j)
= ±2√2 (-1+j)
x^2 = -16j = 16cis(3π/2)
x = ±4 cis(3π/4)
= ±4 (-1/√2 + 1/√2 j)
= ±2√2 (-1+j)
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