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

User Gecko
by
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2 Answers

1 vote

Answer:

The answer is A in edge. The answer is 2.

Explanation:

User Peter Nazarov
by
6.1k points
2 votes
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.
User Wizardzeb
by
6.7k points
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