Modulus and arguments of a complex number

If we represent complex numbers z = a + bi as points on a plane with coordinates (a,b) in a rectangular coordinate system, where the X axis is the axis of real numbers, the Y axis is the axis of imaginary numbers, then a one-to-one correspondence will be established between the points (a,b) in this coordinate system and the complex numbers z = a + bi. Thus, a complex number can also be considered as an ordered pair of real numbers (a;b)=a+bi. It is often convenient to consider a complex number (a;b)=a+bi as a vector with a beginning at the point (0;0) and an end at the point (a;b). The point (0;0) then corresponds to the zero vector.
The correspondence established between the set of complex numbers on the one hand and the set of points or vectors of the plane on the other hand allows the complex numbers a+bi to be called points or vectors. In this case, the operations of addition, subtraction, and multiplication by an integer for complex numbers can be geometrically represented as the corresponding operations for vectors.

Modulus of a complex number

The modulus or absolute value |z| of a complex number z = (a;b) = a+bi is the length of the vector corresponding to this number It is obvious that conjugate complex numbers have equal modules and the following relation is valid for them:: For a real number z = a + 0*i, the modulus coincides with the absolute value of the number a.

Examples.
1) |3-2i| = √(3²+(-2)²) = √(9+4) = √13;
2) |0+i| = |0+1i| = √(0+1²) = √1 = 1;
3) |2√6+5i| = √((2√6)²+5²) = √49 = 7.

Modulus of the difference of complex numbers

By definition of the modulus, the number |z₁-z₂| is the length of the vector of the difference between two vectors z₁ and z₂, but the vector z = z₁-z₂ - is a vector with a start at z₂ and an end at z₁. Thus, the modulus of the difference between two complex numbers is the distance between the points of the complex plane that correspond to these numbers.
Examples.
Which sets of points in the complex plane are given by the conditions:
1) |z-i|=1; 2) |2+z|<|2-z|; 3) 2≤|z-1+2i|<3.
Solution.
1) The condition |z-i|=1 is satisfied by those and only those points of the complex plane that are removed from the point i by a distance equal to one. Such points lie on a circle of unit radius with the center at the point i.

2) Let's rewrite the original inequality as |z-(-2)|<|z-2|. Now the problem can be reformulated as follows: what is the set of points in the complex plane that are closer to the point z=-2 than to the point 2? It is clear that these points are all the points in the plane that lie to the left of the imaginary axis, and only them.

3) Let's rewrite the original inequality 2≤|z-(1-2i)|<3. Complex numbers satisfying this double inequality are removed from the point 1-2i by a distance greater than or equal to two, but less than three. Such points are located inside the ring formed by two concentric circles with the center at the point 1-2i and with radii r₁=2 and r₂=3, and on the inner boundary of the ring.

Complex numbers z, having the same modulus |z| = r, obviously correspond to points of the complex plane located on a circle with center (0;0) and radius r. If |z|≠0, then there are infinitely many complex numbers with this modulus. And there is only one complex number z=0 that has a modulus equal to zero.

Arguments of a complex number

It is geometrically obvious that the complex number z≠0, z=a+bi will be uniquely determined if, in addition to the modulus, we specify, for example, the value of the angle φ between the positive direction of the X axis and the vector z: The argument of the complex number z≠0 is the value of the angle between the positive direction of the X axis and the z vector, where the value of the angle is considered positive if the counting is counterclockwise, and negative if clockwise. Obviously, the argument of a complex number is not uniquely defined, since any angle that differs by a multiple of 2π will also be an argument of the number z, that is, each complex number has an infinite set of arguments.
Thus, specifying the modulus and the argument uniquely defines a complex number. For the number z=0, the argument is not defined, but only in this case the number is set by its modulus. The argument of a complex number, unlike a modulus, is defined ambiguously.
The real and imaginary parts of a complex number z=a+bi, as the figure shows, are expressed through its modulus r=|z| and argument φ as follows:

a=rcosφ; b=rsinφ.

Let's rewrite this system as follows:

cosφ=a/r; sinφ=b/r. ⇔ cosφ=a/√(a²+b²); sinφ=b/√(a²+b²).

Obviously, there are an infinite number of solutions φ to these equations, since any two arguments of a complex number differ by a multiple of 2π. The entire set of arguments is denoted by arg(z) or arg(a+bi). If one particular argument is meant, it is usually denoted by the letter φ.

Examples of finding the arguments of a complex number

Example 1.
Find the arguments of a complex number i.
Solution.
a+bi = i ⇔ a=0; b=1 ⇔ r=√(0+1)=1; cosφ = a/r = 0; sinφ = b/r = 1. These values ​​of cosine and sine correspond to the value of the argument φ=π/2.
⇔ r=1; φ=π/2.
Thus, arg(z) = π/2+2πk, where k is an arbitrary integer.
Answer: π/2+2πk, k∈Z.

Example 2.
Find the arguments of a complex number -1+i.
Solution.
a+bi = -1+i ⇔ a=-1; b=1 ⇔ r=√(1+1)=√2; cosφ = a/r = -1/√2; sinφ = b/r = 1/√2 ⇔ r=√2; φ=3π/4.
Thus, arg(z) = 3π/4+2πk, where k is an arbitrary integer.
Answer: 3π/4+2πk, k∈Z.

Example 3.
Find the arguments of a complex number -1-√3i.
Solution.
a+bi = -1-√3i ⇔ a=-1; b=-√3 ⇔ r=√(1+3)=2; cosφ = a/r = -1/2; sinφ = b/r = -√3/2 ⇔ r=2; φ=-2π/3.
Thus, arg(z) = -2π/3+2πk, where k is an arbitrary integer.
Answer: -2π/3+2πk, k∈Z.

The arguments of a complex number can be found in another way. From the figure it follows that each of the arguments satisfies the equation:

tgφ=b/a.

This equation has more solutions than the system of equations we used earlier. But selecting the right solutions is not difficult, since the algebraic form of the complex number immediately tells us in which quadrant of the complex plane the number is located.
xample 4.
Find the arguments of a complex number -√3+i.
Solution.
a+bi = -√3+i ⇔ a=-√3; b=1 => tgφ = b/a = -1/√3 => φ=-π/6.
But since the number z=-√3+i is located in the second quadrant of the complex plane, its arguments will be numbers φ=5π/6+2πk, k∈Z.
Answer: 5π/6+2πk, k∈Z.