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Naily [24]
4 years ago
5

If the complex number x = 3 + bi and |x|^2 = 13, which is a possible value of b?

Mathematics
1 answer:
Sergio039 [100]4 years ago
6 0
So, if z = x+iy,
|z| = \(\sqrt{x^2+y^2}\)
and \( |z|^2 = x^2 + y^2 \)

so if x = 3+bi
|x|^2 = 3^2 + b^2
which is given to be 13

3^2 + b^2 = 13
9+b^2 = 13
b^2 = 13-9
b^2 = 4
b = +\sqrt 4 , -sqrt 4
b = +2,-2

Hence, the possible values of b are +2 or -2.
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