2 Answers | Add Yours
A complex number has to be found that has a square of 3i.
Let the number be (a + ib)
(a + ib)^2 = 3i
=> a^2 + 2iab + i^2*b^2
=> a^2 - b^2 + 2ab*i
equate the real and imaginary coefficients
2ab = 3
=> a = 3/2*b
substitute in a^2 - b^2 = 0
=> 9/4b^2 = b^2
=> 9 = 4b^4
=> 4b^4 - 9 = 0
=> (2b^2 - 3)(2b^2 + 3) = 0
2b^2 = 3
=> b = sqrt(3/2) and b = -sqrt(3/2)
The other root gives complex values for b and can be ignored.
a = 3/2*b = 3/2*sqrt (3/2) = sqrt (3/2) and a = -sqrt (3/2)
The required complex number is sqrt(3/2) + i*sqrt (3/2) and sqrt -(3/2) - i*sqrt(3/2)
We'll write the rectangular form of the complex number z=a+bi.
From enunciation we find that z^2=3i.
We'll raise to square the rectangular form:
But, from the theory of complex numbers, i^2=-1
Comparing both sides, we'll get:
a^2+b^2=0, a^2=-b^2, so a=b
Also, the imaginary part from the left side has to be equal with the im. part from the right side.
But, as we have demonstrated before, a=b
So, the complex number z, whose square is 3i, is:
We’ve answered 318,913 questions. We can answer yours, too.Ask a question