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Chapter 2: Problem 147

The inequality \(|z-4|<|z-2|\) represent the region given by (a) \(\operatorname{Re}(z)>0\) (b) \(\operatorname{Re}(z)<0\) (c) \(\operatorname{Re}(z)>2\) (d) \(\operatorname{Re}(z)>3\)

Short Answer

Expert verified
The inequality \(|z-4| < |z-2|\) represents the region given by \(\operatorname{Re}(z)<3\).

Step by step solution

01

Understand the problem

Given the inequality \(|z-4| < |z-2|\), it involves the absolute value of the complex number z. The objective is to find the region represented by this inequality in terms of \(\operatorname{Re}(z)\), the real part of the complex number z.
02

Set up the inequality

Write down the given inequality \(|z-4| < |z-2|\). We will now transform this inequality using the properties of absolute values and complex numbers.
03

Expand the inequality

Square both sides of the inequality to remove the absolute value of the complex numbers, and get an algebraic equation: \((z - 4)(z - 4)^* < (z - 2)(z - 2)^*\)
04

Rewrite the inequality in terms of real and imaginary parts

Since z is a complex number, we can rewrite it as \(z = x + yi\), where x and y are real numbers. Replace the occurrences of z in the inequality with \(x + yi\): \(((x + yi) - 4)((x + yi) - 4)^* < ((x + yi) - 2)((x + yi) - 2)^*\)
05

Simplify the inequality

Simplify the inequality step by step: \[((x - 4) + yi)((x - 4) - yi) < ((x - 2) + yi)((x - 2) - yi)\] \[((x - 4)^2 + y^2i^2) < ((x - 2)^2+ y^2i^2)\] \[((x - 4)^2- y^2) < ((x - 2)^2-y^2)\]
06

Solve for Re(z)

To find the region represented by this inequality in terms of Re(z), isolate x in the inequality and then solve for x: \((x-4)^2-y^2<(x-2)^2-y^2\) \((x-4)^2<(x-2)^2\) \((x-4+x-2)(x-4-(x-2))<0\) \((2x-6)(2)<0\) \(x-3<0\) Hence, the inequality \(|z-4| < |z-2|\) represents the region given by \(\operatorname{Re}(z)<3\). The correct option is not provided in the answer choices.

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Most popular questions from this chapter

The area of the triangle in the Arg and diagram formed by the Complex number \(\mathrm{z}\), iz and \(\mathrm{z}+\mathrm{iz}\) is (a) \(|z|^{2}\) (b) \((\sqrt{3} / 2)|z|^{2}\) (c) \((1 / 2)|\mathrm{z}|^{2}\) (d) \((3 / 2)|\mathrm{z}|^{2}\)

Evaluate \(\left[\mathrm{i}^{19}+(1 / \mathrm{i})^{25}\right]^{2}\) (a) 4 (b) \(-4\) (c) 5 (d) \(-5\)

If \(\mathrm{z}_{1}=2-\mathrm{i}\) and \(\mathrm{z}_{2}=1+\mathrm{i}\) then \(\left|\left[\left(z_{1}-z_{2}+1\right) /\left(z_{1}+z_{2}+i\right)\right]\right|=\) (a) \(\sqrt{(5 / 3)}\) (b) \(\sqrt{(3 / 5)}\) (c) \(\sqrt{(4 / 5)}\) (d) \(\sqrt{(5 / 4)}\)

If \([\\{(1+i) x-2 i\\} /(3+i)]+[\\{(2-3 i) y+i\\} /(3-i)\\}]=i\) then \((\mathrm{x}, \mathrm{y})=\) (a) \((3,1)\) (b) \((3,-1)\) (c) \((-3,1)\) (d) \((-3,-1)\)

\(w \neq 1\) is cube root of 1 and \(\left(1+w^{2}\right)^{n}=\left(1+w^{4}\right)^{n}\) then the smallest positive value of \(\mathrm{n}\) is (a) 2 (b) 3 (c) 5 (d) 6
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