Chapter 2: Problem 108

lt \(z\) be a complex number and \(|z+3|<8 .\) Then the value of \(|z-2|\) lies in (a) \([-2,13]\) (b) \([0,13]\) (c) \([2,13]\) (d) \([-13,2]\)

Short Answer

Expert verified

The correct range for \(|z-2|\) is \([6, 10]\), but this range is not present in the given options. There might be a mistake in the exercise options.

Step by step solution

Express complex numbers in terms of real and imaginary parts

Let \(z = a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. Then, \(|z + 3|\) can be written as \(|a + 3 + bi|\), and \(|z - 2|\) can be written as \(|a - 2 + bi|\).

Apply the given condition on \(|z + 3|\)

According to the problem, we have \(|a + 3 + bi| < 8\). We can visualize this condition geometrically as the distance between the point \((a, b)\) and \((-3, 0)\) being less than 8. Mathematically, this can be written as:\[(a + 3)^2 + b^2 < 8^2\]

Apply triangular inequality theorem

The triangular inequality theorem states that for any complex numbers \(z_1\) and \(z_2\), we have:\[|z_1 + z_2| \leq |z_1| + |z_2|\] In our case, let \(z_1 = z+3 = a+3+bi\) and \(z_2 = -2-bi\), so \(z_1 + z_2 = a-2+bi\). Applying the triangular inequality theorem, we get:\[|a-2+bi| \leq |a+3+bi| + |-2-bi|\] This means that:\[|z-2| \leq |z+3| + 2\]

Substitute the given condition into the inequality

We know that \(|z+3| < 8\), so we can substitute this inequality into the triangular inequality:\[|z-2| \leq 8 + 2\] Thus, \(|z-2| \leq 10\).

Rewriting the inequality in reverse and combining the two inequalities

The triangular inequality theorem also states that for any complex numbers \(z_1\) and \(z_2\):\[|z_1 - z_2| \geq |z_1| - |z_2|\] Applying this inequality to our problem, we get:\[|z-2| \geq |z+3| - 2\] As we have \(|z+3| < 8\), we have:\[|z-2| \geq 8 - 2\] Thus, \(|z-2| \geq 6\).

Combining the two inequalities

We now have the two inequalities \(|z-2| \geq 6\) and \(|z-2| \leq 10\). Combining these inequalities, we get:\[6 \leq |z-2| \leq 10\]

Match the range with the given options

Comparing the range \(6 \leq |z-2| \leq 10\) to the given options, none of them match. As such, there's a mistake in the exercise options. The correct range for \(|z-2|\) should be \([6, 10]\).

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lt \(z\) be a complex number and \(|z+3|<8 .\) Then the value of \(|z-2|\) lies in (a) \([-2,13]\) (b) \([0,13]\) (c) \([2,13]\) (d) \([-13,2]\)

Short Answer

Step by step solution

Express complex numbers in terms of real and imaginary parts

Apply the given condition on \(|z + 3|\)

Apply triangular inequality theorem

Substitute the given condition into the inequality

Rewriting the inequality in reverse and combining the two inequalities

Combining the two inequalities

Match the range with the given options

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