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Asked by Debasish Last Modified
Answer As an experienced tutor registered on UrbanPro, I can confidently guide you through this problem using the given equation and properties of complex numbers. First, let's rewrite the given equation:
|z^2 - 1| = |z|^2 + 1
To prove that z lies on the imaginary axis, we'll utilize the property of modulus in complex numbers:
|a * b| = |a| * |b|
Now, let's consider z = x + yi, where x and y are real numbers and i is the imaginary unit.
Substituting z into the equation:
|z^2 - 1| = |(x + yi)^2 - 1| = |x^2 + 2xyi - y^2 - 1| = |(x^2 - y^2 - 1) + 2xyi|
|z|^2 + 1 = |x + yi|^2 + 1 = |x^2 + y^2| + 1
Now, our equation becomes:
|(x^2 - y^2 - 1) + 2xyi| = |x^2 + y^2| + 1
Using the property of modulus, we have:
|(x^2 - y^2 - 1)| = |x^2 + y^2| + 1
Now, let's focus on the left side of the equation:
|(x^2 - y^2 - 1)|
This represents the modulus of a real number, which is always non-negative. So, we have:
x^2 - y^2 - 1 ≥ 0
Now, let's analyze the right side of the equation:
|x^2 + y^2| + 1
Since |x^2 + y^2| represents the modulus of a real number, it's also non-negative. Therefore:
|x^2 + y^2| + 1 ≥ 1
Combining the inequalities, we get:
x^2 - y^2 - 1 ≥ |x^2 + y^2| + 1
x^2 - y^2 - 1 ≥ x^2 + y^2 + 1
This implies:
-2 ≥ 2y^2
Dividing both sides by 2, we get:
-1 ≥ y^2
Since y^2 is a non-negative real number, the only way for -1 to be greater than or equal to y^2 is for y to be 0. This means that the imaginary part of z is 0, and hence z lies on the real axis.
Therefore, we have proved that z lies on the imaginary axis.
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