In mathematics, the set of complex number is created as an extension of the set of real numbers, containing in particular an imaginary number noted ia, b such that i2 = −1. The square of (−i) is also equal to −1: (−i) 2 = −1.
Any complex number can be written as a + i b where a and b are real numbers.
We can provide the set of complex numbers with an addition and a multiplication which make it a commutative field containing the field of real numbers. It is called the field of complex numbers and is written ℂ. The notion of absolute value defined on the set of real numbers can be extended to the set of complex numbers and then takes the name of module. But we cannot provide the set of complex numbers with an order relation that would make it into a totally ordered field, that is to say that it is not possible to compare two complexes while respecting the rules operating procedures valid for real numbers.
Very simple, add up the real parts (without i) and add up the imaginary parts (with i):
This is equal to use rule: (a+bi)+(c+di) = (a+c) + (b+d)i
(1+i) + (6-5i) = 7-4i
12 + 6-5i = 18-5i
(10-5i) + (-5+5i) = 5
Again very simple, subtract the real parts and subtract the imaginary parts (with i):
This is equal to use rule: (a+bi)+(c+di) = (a-c) + (b-d)i
(1+i) – (3-5i) = -2+6i
-1/2 – (6-5i) = -6.5+5i
(10-5i) – (-5+5i) = 15-10i
To multiply two complex numbers, use distributive law, avoid binomials, and apply i2 = -1.
This is equal to use rule: (a+bi)(c+di) = (ac-bd) + (ad+bc)i
(1+i) (3+5i) = 1*3+1*5i+i*3+i*5i = 3+5i+3i-5 = -2+8i
-1/2 * (6-5i) = -3+2.5i
(10-5i) * (-5+5i) = -25+75i
The division of two complex numbers can be accomplished by multiplying the numerator and denominator by the complex conjugate of the denominator. This avoid imaginary unit i from the denominator. If the denominator is c+di, to make it without i (or make it real), just multiply with conjugate c-di:
(c+di)(c-di) = c2+d2
(10-5i) / (1+i) = 2.5-7.5i
-3 / (2-i) = -1.2-0.6i
6i / (4+3i) = 0.72+0.96i
Absolute value or modulus
The absolute value or modulus is the distance of the image of a complex number from the origin in the plane. The calculator uses the Pythagorean theorem to find this distance. Very simple, see examples: |3+4i| = 5
|1-i| = 1.4142136
|6i| = 6
abs(2+5i) = 5.3851648
Square root of complex number (a+bi) is z, if z2 = (a+bi). Here ends simplicity. Because of the fundamental theorem of algebra, you will always have two different square roots for a given number. If you want to find out the possible values, the easiest way is probably to go with De Moivre’s formula. Here our calculator is on edge, because square root is not a well defined function on complex number. We calculate all complex roots from any number – even in expressions:
sqrt(9i) = 2.1213203+2.1213203i
sqrt(10-6i) = 3.2910412-0.9115656i
pow(-32,1/5)/5 = -0.4
pow(1+2i,1/3)*sqrt(4) = 2.439233+0.9434225i
pow(-5i,1/8)*pow(8,1/3) = 2.3986959-0.4771303i
Square, power, complex exponentiation
Our calculator can power any complex number to any integer (positive, negative), real, or even complex number. In other words, we calculate ‘complex number to a complex power’ or ‘complex number raised to a power’…
i^2 = -1
i^61 = i
(6-2i)^6 = -22528-59904i
(6-i)^4.5 = 2486.1377428-2284.5557378i
(6-5i)^(-3+32i) = 2929449.03994-9022199.58262i
i^i = 0.2078795764
pow(1+i,3) = -2+2i
- Square Root of a value or expression.
- the sine of a value or expression. Autodetect radians/degrees.
- the cosine of a value or expression. Autodetect radians/degrees.
- tangent of a value or expression. Autodetect radians/degrees.
- e (the Euler Constant) raised to the power of a value or expression
- Power one complex number to another integer/real/complex number
- The natural logarithm of a value or expression
- The base-10 logarithm of a value or expression
- abs or |1+i|
- The absolute value of a value or expression
- Phase (angle) of a complex number
- is less known notation: cis(x) = cos(x)+ i sin(x); example: cis (pi/2) + 3 = 3+i
- conjugate of complex number – example: conj(4i+5) = 5-4i
Complex Number Exercises
1. Evaluate the following expressions:
a) (3 + 2i) – (8 – 5i)
b) (4 – 2i)*(1 – 5i)
c) (- 2 – 4i) / i
d) (- 3 + 2i) / (3 – 6i)
a) -5 + 7i
b) -6 – 22i
c) -4 + 2i
d) -7/15 – 4i/15
2. Find all complex numbers of the form z = a + bi , where a and b are real numbers such that z z’ = 25 and a + b = 7
where z’ is the complex conjugate of z.
z z’ = (a + bi)(a – bi)
= a2 + b2 = 25
a + b = 7 gives b = 7 – a
Substitute above in the equation a2 + b2 = 25
a2 + (7 – a)2 = 25
Solve the above quadratic function for a and use b = 7 – a to find b.
a = 4 and b = 3 or a = 3 and b = 4
z = 4 + 3i and z = 3 + 4i have the property z z’ = 25.
3. If (x + yi) / i = ( 7 + 9i ) , where x and y are real, what is the value of (x + yi)(x – yi)?
(x + yi) / i = ( 7 + 9i )
(x + yi) = i(7 + 9i) = -9 + 7i
(x + yi)(x – yi) = (-9 + 7i)(-9 – 7i) = 81 + 49 = 130
4. Calculate: z = (10-2i) + (-5+235i)
z = 5+233i
Angle notation (phasor):
z = 233.0536419 ∠ 88°46’14″
z = 233.0536419 × (cos 88°46’14″ + i sin 88°46’14″)
z = 233.0536419 × ei 0.4931704
r = |z| = 233.0536419 … magnitude (modulus, absolute value)
θ = arg z = 1.5493404 rad = 88.77067° = 88°46’14″ = 0.4931704π rad … angle (argument or phase)
Cartesian form of imaginary number: z = 5+233i
Real part: x = Re z = 5
Imaginary part: y = Im z = 233
5. Find all complex numbers z such that (4 + 2i)z + (8 – 2i)z’ = -2 + 10i, where z’ is the complex conjugate of z.
Let z = a + bi where a and b are real numbers. The complex conjugate z’ is written in terms of a and b as follows: z’= a – bi. Substitute z and z’ in the given equation
(4 + 2i)(a + bi) + (8 – 2i)(a – bi) = -2 + 10i
Expand and separate real and imaginary parts.
(4a – 2b + 8a – 2b) + (4b + 2a – 8b – 2a )i = -2 + 10i
Two complex numbers are equal if their real parts and imaginary parts are equal. Group like terms.
12a – 4b = -2 and – 4b = 10
Solve the system of the unknown a and b to find:
b = -5/2 and a = -1
z = -1 – (5/2)i
6. P(z) = z4 + a z3 + b z2 + c z + d is a polynomial where a, b, c and d are real numbers. Find a, b, c and d if two zeros of polynomial P are the following complex numbers: 2 – i and 1 – i.
Since all coefficients of polynomial P are real, the complex conjugate to the given zeros are also zeros of P. Hence
P(z) = (z – (2 – i))(z – (2 + i))(z – (1 – i))(z – (1 + i)) =
= z4 – 6 z3 + 15 z2 – 18 z + 10
Hence: a = -6, b = 15, c = -18 and d = 10.
7. Determine all complex number z that satisfy the equation: z + 3 z’ = 5 – 6i
where z’ is the complex conjugate of z.
Let z = a + bi , z’ = a – bi ; a and b real numbers.
Substituting z and z’ in the given equation obtain
a + bi + 3*(a – bi) = 5 – 6i
a + 3a + (b – 3b) i = 5 – 6i
4a = 5 and -2b = -6
a = 5/4 and b = 3
z = 5/4 + 3i
8. The complex number 2 + 4i is one of the root to the quadratic equation x2 + bx + c = 0, where b and c are real numbers.
a) Find b and c
b) Write down the second root and check it.
a) Substitute solution in equation: (2 + 4i)2 + b(2 + 4i) + c = 0
Expand terms in equation and rewrite as: (-12 + 2b + c) + (16 + 4b)i = 0
Real part and imaginary part equal zero.
-12 + 2b + c = 0 and 16 + 4b = 0
Solve for b: b = -4 , substitute and solve for c: c = 20
b) Since the given equation has real numbers, the second root is the complex conjugate of the given root: 2 – 4i is the second solution.
Check: (2 – 4i)2 – 4 (2 – 4i) + 20
(Expand) = 4 – 16 – 16i – 8 + 16i + 20
= (4 – 16 – 8 + 20) + (-16 + 16)i = 0
9. Exercises complex number: a) Show that the complex number 2i is a root of the equation: z4 + z3 + 2 z2 + 4 z – 8 = 0
b) Find all the roots root of this equation.
a) (2i)4 + (2i)3 + 2 (2i)2 + 4 (2i) – 8
= 16 – 8i – 8 + 8i – 8 = 0
b) 2i is a root -2i is also a root (complex conjugate because all coefficients are real).
z4 + z3 + 2 z2 + 4 z – 8 = (z – 2i)(z + 2i) q(z)
= (z2 + 4)q(z)
q(z) = z2 + z – 2
The other two roots of the equation are the roots of q(z): z = 1 and z = -2.
10. Calculate: z = (10-2i) / (-5+25i)
- Complex number: 10-2i
- Complex number: -5+25i
- Divide: the result of step No. 1 / the result of step No. 2 = (10-2i) / (-5+25i) = 10-2i/ = (10-2i)*(-5-25i)/ = = = = = -100-240i/ = -0.1538462-0.3692308iTo divide complex numbers, you must multiply both (numerator and denominator) by the conjugate of the denominator. To find the conjugate of a complex number, you change the sign in imaginary part.
Distribute in both the numerator and denominator to remove the parenthesis and add and simplify.
z = -0.1538462-0.3692308i
Angle notation (phasor):
z = 0.4 ∠ -112°37’11″
z = 0.4 × (cos (-112°37’11″) + i sin (-112°37’11″))
z = 0.4 × ei (-0.6256659)
r = |z| = 0.4 … magnitude (modulus, absolute value)
θ = arg z = -1.9655874 rad = -112.61986° = -112°37’11″ = -0.6256659π rad … angle (argument or phase)
Cartesian form of imaginary number: z = -0.1538462-0.3692308i
Real part: x = Re z = -0.154
Imaginary part: y = Im z = -0.36923077
11. Find all complex numbers z such that z2 = -1 + 2 sqrt(6) i.
Let z = a + bi
Substitute into given equation: (a + bi)2 = -1 + 2 sqrt(6) i
Expand: a2 – b2 + 2 ab i = – 1 + 2 sqrt(6) i
Real part and imaginary parts must be equal.
a2 – b2 = – 1 and 2 ab = 2 sqrt(6)
Equation 2 ab = 2 sqrt(6) gives: b = sqrt(6) / a
Substitute: a2 – ( sqrt(6) / a )2) = – 1
a4 – 6 = – a2
Solve above equation and select only real roots: a = sqrt(2) and a = – sqrt(2)
Substitute to find b and write the two complex numbers that satisfies the given equation.
z1 = sqrt(2) + sqrt(3) i , z2 = – sqrt(2) – sqrt(3) i
12. Given that the complex number z = -2 + 7i is a root to the equation: z3 + 6 z2 + 61 z + 106 = 0
find the real root to the equation.
Since z = -2 + 7i is a root to the equation and all the coefficients in the terms of the equation are real numbers, then z’ the complex conjugate of z is also a solution. Hence
z3 + 6 z2 + 61 z + 106 = (z – (-2 + 7i))(z – (-2 – 7i)) q(z)
= (z2 + 4z + 53) q(z)
q(z) = [ z3 + 6 z2 + 61 z + 106 ] / [ z2 + 4z + 53 ] = z + 2
Z + 2 is a factor of z3 + 6 z2 + 61 z + 106 and therefore z = -2 is the real root of the given equation.