The inequality $|z-4| 3$.

The inequality |z-4|<|z-2| represents the region given by x>3.

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|z-4|<|z-2|

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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The inequality $|z-4|<|z-2|$ means that the distance from the complex number $z$ to $4$ is less than its distance to $2$ . Geometrically, this describes all points $z$ that lie to the right of the perpendicular bisector of the segment joining $2$ and $4$ , which is the line $x=3$ . Thus, the region is $x>3$ .

Understanding inequalities involving the modulus of complex numbers is fundamental. The key concept here is the geometric interpretation of the modulus.

The expression $|z_1 - z_2|$ represents the distance between the complex numbers $z_1$ and $z_2$ in the Argand plane.

In our problem:

$|z-4|$ represents the distance between the complex number $z$ and the complex number $4$ (which can be written as $4+0i$ ).
$|z-2|$ represents the distance between the complex number $z$ and the complex number $2$ (which is $2+0i$ ).

So, the inequality $|z-4| < |z-2|$ asks for all complex numbers $z$ that are closer to $4$ than they are to $2$ .

Consider the two fixed points $A=(2,0)$ and $B=(4,0)$ in the complex plane. We are looking for points $P=(x,y)$ such that the distance $PB$ is less than the distance $PA$ .

The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining those points.

The midpoint of the segment joining $(2,0)$ and $(4,0)$ is $\left(\frac{2+4}{2}, \frac{0+0}{2}\right) = (3,0)$ .

Since the segment joining $(2,0)$ and $(4,0)$ is horizontal, its perpendicular bisector is a vertical line passing through the midpoint $(3,0)$ . This line is $x=3$ .

Points closer to $(4,0)$ than to $(2,0)$ must lie on the side of the line $x=3$ that contains $(4,0)$ . Since $(4,0)$ is to the right of $x=3$ , the region is $x>3$ .

Let's confirm this intuition with an algebraic approach.

Represent $z$ in Cartesian form:

Let $z = x+iy$ , where $x$ and $y$ are real numbers.
Substitute $z$ into the inequality:

The given inequality is $|z-4| < |z-2|$ .

Substitute $z=x+iy$ :

$|(x+iy)-4| < |(x+iy)-2|$

Group the real and imaginary parts:

$|(x-4)+iy| < |(x-2)+iy|$
Apply the definition of modulus:

For a complex number $a+bi$ , its modulus is $|a+bi| = \sqrt{a^2+b^2}$ . …

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Q.The inequality ∣z−4∣<∣z−2∣|z-4|<|z-2|∣z−4∣<∣z−2∣ represents the region given by x>3x>3x>3.

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Q.The inequality $|z-4|<|z-2|$ represents the region given by $x>3$ .

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