If $|z+4|\leq3$, then the greatest and least values of $|z+1|$ are _____ and _____.

If |z+4|≤3, then the greatest and least values of |z+1| are ___ and ___.

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|z+4|\leq3

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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 $|z+4|\leq 3$ describes a closed disk in the complex plane centered at $-4$ with radius $3$ . The extreme values of $|z+1|$ occur at points on this disk that are farthest from and closest to $-1$ , giving maximum $6$ and minimum $0$ .

The key insight here is geometric. When we work with moduli of complex numbers, we're really measuring distances in the plane. The condition $|z+4|\leq 3$ tells us that $z$ lies within or on a circle, and we need to find how far points in this region can be from another fixed point.

Let me rewrite the expressions to make the geometry transparent. The inequality $|z+4|\leq 3$ is equivalent to $|z-(-4)|\leq 3$ , which means $z$ lies in a closed disk of radius $3$ centered at the point $-4$ on the real axis. Similarly, $|z+1| = |z-(-1)|$ measures the distance from $z$ to the point $-1$ .

So the question becomes: among all points in the disk centered at $-4$ with radius $3$ , which are farthest from and closest to $-1$ ?

The distance between the two centers is $|-1-(-4)| = |3| = 3$ . This is exactly equal to the radius of the disk, which means the point $-1$ lies on the boundary of the disk. This is a special case worth noting.

Finding the maximum value:

The farthest point from $-1$ in the disk will lie on the boundary (the circle $|z+4|=3$ ), on the ray from $-1$ through $-4$ extended outward.
Moving from $-1$ toward $-4$ , we travel a distance of $3$ to reach the center. Then continuing in the same direction for another $3$ units (the radius) takes us to the far edge of the disk.
The maximum distance is therefore $3 + 3 = 6$ .

Algebraically, the farthest point is $z = -4 - 3 = -7$ , and indeed $|-7+1| = |-6| = 6$ .

Finding the minimum value:

The closest point from $-1$ in the disk will also lie on the boundary, but in the opposite direction—on the ray from $-1$ toward $-4$ . …

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Q.If ∣z+4∣≤3|z+4|\leq3∣z+4∣≤3, then the greatest and least values of ∣z+1∣|z+1|∣z+1∣ are _____ and _____.

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Q.If $|z+4|\leq3$ , then the greatest and least values of $|z+1|$ are _ and _.

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