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Chapter 2: Problem 151

If \(z\) is a complex number satisfying \(|z-\operatorname{iRe}(z)|=|z-\operatorname{Im}(z)|\) then \(\mathrm{z}\) lies on (a) \(\mathrm{y}=\mathrm{x}-1\) (b) \(\mathrm{y}=\pm \mathrm{x}\) (c) \(\mathrm{y}=\mathrm{x}+1\) (d) \(\mathrm{y}=-\mathrm{x}+1\)

Short Answer

Expert verified
In this problem, we are given the equation \(|z-\operatorname{iRe}(z)|=|z-\operatorname{Im}(z)|\), where \(z\) is a complex number, and we need to find a real equation that represents the set of all complex numbers \(z\) that satisfy this equation. By substituting \(z = x + yi\), where \(x = \operatorname{Re}(z)\) and \(y = \operatorname{Im}(z)\), and applying properties of moduli, we can derive the equation \(y^2 = x^2\), which leads to the solution \(y = \pm x\). Therefore, the complex numbers \(z\) that satisfy the given equation lie on the lines \(y=x\) and \(y=-x\). The correct answer is (b) \(\mathrm{y}=\pm \mathrm{x}\).

Step by step solution

01

Write z as x + yi

Let \(z=x+\mathrm{i}y\), where \(x=\operatorname{Re}(z)\) and \(y=\operatorname{Im}(z)\). We will substitute this in the given equation.
02

Substitute z with x + yi in the equation

We can rewrite the equation as follows: \(|x+\mathrm{i}y-\mathrm{i}(x)|=|x+\mathrm{i}y-(y)|\)
03

Simplify the equation

Now let's simplify the equation and see if we can make it more manageable: \(|x+\mathrm{i}(y-x)|=|x-y+\mathrm{i}y|\)
04

Use the property of moduli

We can use the property that the modulus of a complex number \(a+\mathrm{i}b\) is \(\sqrt{a^2+b^2}\). Therefore, our equation becomes: \(\sqrt{(x^2+(y-x)^2)}=\sqrt{((x-y)^2+y^2)}\)
05

Square both sides and rearrange

We can get rid of the square root by squaring both sides of the equation. Then, we can rearrange the terms of the equation to solve for \(y\) in terms of \(x\): \[(x^2 + (y-x)^2) = ((x-y)^2 + y^2) \] Rearrange and simplify to get: \[x^2 - 2x^2 + x^2 + y^2 = x^2 + 2x^2 - y^2\]
06

Solve for y

To find the relationship between \(x\) and \(y\), we'll solve for \(y\) in terms of \(x\): \[y^2 = x^2\] \[y = \pm x\] So, the complex numbers \(z\) that satisfy the given equation lie on the lines \(y=x\) and \(y=-x\). The correct answer is then (b) \(\mathrm{y}=\pm \mathrm{x}\)

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Most popular questions from this chapter

Let \(\alpha, \beta\) be real and \(z\) be a complex number if \(z^{2}+\alpha z+\beta=0\) has two distinct roots on the line \(\operatorname{Re}(z)=1\) then it is necessary that (a) \(\beta \in(-1,0)\) (b) \(|\beta|=1\) (c) \(\beta \in[1, \infty)\) (d) \(\beta \in(0,1)\)

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) \([-13,2]\)

If \(\mathrm{w}=[\mathrm{z} /\\{\mathrm{z}-(1 / 3)\\}] \mathrm{i}\) and \(|\mathrm{w}|=1\) then \(\mathrm{z}\) lies on (a) circle (b) an ellipse (c) Parabola (d) a straight line

\(w \neq 1\) 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

If \(|z|=1\) and \(w=[(z-1) /(z+1)](z \neq-1)\) then \(\operatorname{Re}(w)=\) (a) 0 (b) \(\left[1 /\left(|z+1|^{2}\right)\right]\) (c) \(\left[1 /\left(|z+1|^{3}\right)\right]\) (d) \(\left[\sqrt{2} /\left(|\mathrm{z}+1|^{2}\right)\right]\)
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