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Operations with complex numbers: Do you know how to find the square root of a negative number? We know that no real number when squared gives a negative number. This is achieved by the use of complex numbers. Any number of the form \(a + ib\) is called a complex number, where \(a\) and \(b\) are real numbers. For example, \(2 – 14i\) is a complex number, where \(2\) is the real number, \( – 14i\) is the imaginary number, and \(i\) is the imaginary unit.
Like real numbers, the four arithmetic operations can also be performed on complex numbers. These calculations that use complex numbers play a significant role in the fields of quantum mechanics, electrical circuit analysis, and fluid dynamics. The arithmetic operations are:
In this article, let us learn to perform these operations with complex numbers and their various properties.
What do you think happens when two complex numbers are added or subtracted?
They behave similarly to real numbers. The difference here is that since complex numbers have two parts, real and imaginary, they are operated on separately.
Let \({z_1} = a + ib\) and \({z_2} = c + id\) be two complex numbers. Then their sum and difference is written as,
Follow these steps to add or subtract two complex numbers in polar form:
Step 1: Convert the complex numbers from the given polar form to rectangular form.
Step 2: Add or subtract the complex numbers in rectangular form, as described above.
Step 3: Convert the resulting complex number to polar form again. This is the required answer.
| Name of the Property | Definition | Proof/Example | Result |
| Closure Property | A set is closed for some arithmetic operation | \(\left( {4 + i2} \right) + \left( {6 + i3} \right)\) \( = \left( {{\rm{4 + 6}}} \right) + \left( {i2 + i3} \right)\) \( = 10 + i5\) | TRUE |
| Commutative Property | The order of numbers does not alter the result | \((6 + i3) + (4 + i2)\) \( = (6 + 4) + (i3 + i2)\) \( = 10 + i5\) | TRUE |
| Associative Property | Rearranging the parenthesis in an expression will not alter the result | \({z_1} + \left( {{z_2} + {z_3}} \right) = \left( {{\alpha _1} + i{\beta _1}} \right) + \left\{ {\left( {{\alpha _2} + i{\beta _2}} \right) + \left( {{\alpha _3} + i{\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} + i{\beta _1}} \right) + \left\{ {\left( {{\alpha _2} + {\alpha _3}} \right) + i\left( {{\beta _2} + {\beta _3}} \right)} \right\}\) \( = \left\{ {{\alpha _1} + \left( {{\alpha _2} + {\alpha _3}} \right)} \right\} + \left\{ {i\left( {{\beta _1} + {\beta _2} + {\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} + {\alpha _2} + {\alpha _3}} \right) + i\left( {{\beta _1} + {\beta _2} + {\beta _3}} \right)\) \( = \{ {\alpha _1} + \left( {{\alpha _2} + {\alpha _3}} \right)\} + \{ i\left( {{\beta _1} + {\beta _2} + {\beta _3}} \right)\} \) \( = \left( {\left( {{\alpha _1} + {\alpha _2}} \right) + {\alpha _3}} \right) + i\left( {\left( {{\beta _1} + {\beta _2}} \right) + {\beta _3}} \right)\) \( = \left( {\left( {{\alpha _1} + {\alpha _2}} \right) + i\left( {{\beta _1} + {\beta _2}} \right)} \right) + \left( {{\alpha _3} + i{\beta _3}} \right)\) \( = \left( {{z_1} + {z_2}} \right) + {z_3}\) | TRUE |
| Additive Inverse | A number, when added to the given number, gives zero | \({z_1} + A \cdot {\mathop{\rm In}\nolimits} . = 0\) \(A \cdot {\mathop{\rm In}\nolimits} . = 0 – {z_1}\) \(A \cdot {\mathop{\rm In}\nolimits} . = – {z_1}\) | Additive inverse of \({z_1}\) is \( – {z_1}\) |
| Additive Identity | A number, when added to the number, does not alter it | \({z_1} + A \cdot Id. = {z_1}\) \(A \cdot Id. = {z_1} – {z_1}\) \(A \cdot Id. = 0\) | Additive Identity of \({z_1}\) is \(0\) |
| Name of the Property | Definition | Proof/Example | Result |
| Closure Property | A set is closed for some arithmetic operation | \((4 + i2) – (6 + i3)\) \( = (4 – 6) + (i2 – i3)\) \( = – 2 – i\) | TRUE |
| Commutative Property | The order of numbers does not alter the result | \((6 + i3) – (4 + i2)\) \( = (6 – 4) + (i3 – i2)\) \( = 2 + i\) | FALSE |
| Associative Property | Rearranging the parenthesis in an expression will not alter the result | \({z_1} – \left( {{z_2} – {z_3}} \right) = \left( {{\alpha _1} + i{\beta _1}} \right) – \left\{ {\left( {{\alpha _2} + i{\beta _2}} \right) – \left( {{\alpha _3} + i{\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} + i{\beta _1}} \right) – \left\{ {\left( {{\alpha _2} – {\alpha _3}} \right) + i\left( {{\beta _2} – {\beta _3}} \right)} \right\}\) \( = \left\{ {{\alpha _1} + \left( { – {\alpha _2} + {\alpha _3}} \right)} \right\} + \left\{ {i\left( {{\beta _1} – {\beta _2} + {\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} – {\alpha _2} + {\alpha _3}} \right) + i\left( {{\beta _1} – {\beta _2} + {\beta _3}} \right)\) \( = \left( {\left( {{\alpha _1} – {\alpha _2}} \right) + {\alpha _3}} \right) + i\left( {\left( {{\beta _1} – {\beta _2}} \right) + {\beta _3}} \right)\) \( = \left( {\left( {{\alpha _1} – {\alpha _2}} \right) + i\left( {{\beta _1} – {\beta _2}} \right)} \right) + \left( {{\alpha _3} + i{\beta _3}} \right)\) \( = \left( {{z_1} – {z_2}} \right) + {z_3}\) | FALSE |
The process of multiplying two complex numbers is more complicated than the multiplication of real numbers. As there are two parts to a complex number, we find the product of four different pairs of numbers as shown below when multiplying two of them.
For two complex numbers \({z_1} = a + ib\) and \({z_2} = c + id\), their product is given by,
\({z_1}{z_2} = (a + ib)(c + id)\)
\( = ac + iad + ibc + {i^2}bd\)
\({z_1}{z_2} = (ac – bd) + i(ad + bc)\)
Note that \({i^2} = – 1\)
For any two complex numbers, \({z_1} = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\) and \({z_2} = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\), the formula for multiplication in polar form is given by,
| Name of the Property | Proof/Example | Result |
| Closure Property | \((3 + i2) \times (4 + i5)\) \( = 3 \cdot 4 + 3 \cdot i5 + i2 \cdot 4 + i5 \cdot i2\) \( = 12 + i15 + i8 + {i^2}2\) \( = 12 – 2 + i23\) \( = 10 + i23\) | TRUE |
| Commutative Property | \((4 + i5) \times (3 + i2)\) \( = 12 + i15 + i8 + {i^2}2\) \( = 12 – 2 + i23\) \( = 10 + i23\) | TRUE |
| Associative Property | \({z_1} \times \left( {{z_2} \times {z_3}} \right) = \left( {{\alpha _1} + i{\beta _1}} \right) \times \left\{ {\left( {{\alpha _2} + i{\beta _2}} \right) + \left( {{\alpha _3} + i{\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} + i{\beta _1}} \right) \times \left\{ {\left( {{\alpha _2}{\alpha _3} – {\beta _2}{\beta _3}} \right) + i\left( {{\beta _3}{\alpha _2} + {\beta _2}{\alpha _3}} \right)} \right\}\) \( = \left( {{\alpha _1}\left( {{\alpha _2}{\alpha _3} – {\beta _2}{\beta _3}} \right) – {\beta _1}\left( {{\beta _3}{\alpha _2} + {\beta _2}{\alpha _3}} \right)} \right) + i\left\{ {{\alpha _1}\left( {{\beta _3}{\alpha _2} + {\beta _2}{\alpha _3}} \right) + \left( {{\alpha _2}{\alpha _3} – {\beta _2}{\beta _3}} \right){\beta _1}} \right\}\) \( = \left( {{\alpha _1}{\alpha _2}{\alpha _3} – {\alpha _1}{\beta _2}{\beta _3} – {\beta _1}{\beta _3}{\alpha _2} – {\beta _1}{\beta _2}{\alpha _3}} \right) + i\left\{ {{\beta _3}{\alpha _1}{\alpha _2} + {\alpha _1}{\beta _2}{\alpha _3} + {\beta _1}{\alpha _2}{\alpha _3} – {\beta _1}{\beta _2}{\beta _3}} \right\}\) \( = \left\{ {\left( {{\alpha _1}{\alpha _2} – {\beta _1}{\beta _2}} \right){\alpha _3} – \left( {{\beta _1}{\alpha _2} + {\alpha _1}{\beta _2}} \right){\beta _3}} \right\} + i\left\{ {\left( {{\alpha _1}{\alpha _2} – {\beta _1}{\beta _2}} \right){\alpha _3} + \left( {{\beta _1}{\alpha _2} + {\alpha _1}{\beta _2}} \right){\beta _3}} \right\}\) \( = \left\{ {\left( {{\alpha _1}{\alpha _2} – {\beta _1}{\beta _2}} \right) + i\left( {{\beta _1}{\alpha _2} + {\alpha _1}{\beta _2}} \right)} \right\} \times \left( {{\alpha _3} + i{\beta _3}} \right)\) \( = \left\{ {\left( {{\alpha _1} + i{\beta _1}} \right) \times \left( {{\alpha _2} + i{\beta _2}} \right)} \right\} \times \left( {{\alpha _3} + i{\beta _3}} \right)\) \( = \left( {{z_1} \times {z_2}} \right) \times {z_3}\) | TRUE |
| Distributive Property | \({z_1} \times \left( {{z_2} + {z_3}} \right) = \left( {{\alpha _1} + i{\beta _1}} \right) \times \left\{ {\left( {{\alpha _2} + i{\beta _2}} \right) + \left( {{\alpha _3} + i{\beta _3}} \right)} \right\}\) \( = \left( {{\alpha _1} + i{\beta _1}} \right) \times \left\{ {\left( {{\alpha _2} + {\alpha _3}} \right) + i\left( {{\beta _2} + {\beta _3}} \right)} \right\}\) \( = \left\{ {\left( {{\alpha _1} + i{\beta _1}} \right)\left( {{\alpha _2} + {\alpha _3}} \right) + i\left( {{\alpha _1} + i{\beta _1}} \right)\left( {{\beta _2} + {\beta _3}} \right)} \right\}\) \( = \left\{ {\left( {\left( {{\alpha _1}{\alpha _2} + i{\alpha _2}{\beta _1}} \right) + \left( {{\alpha _1}{\alpha _3} + i{\alpha _3}{\beta _1}} \right)} \right) + i\left( {\left( {{\beta _2}{\alpha _1} + {\beta _3}{\alpha _1}} \right) + i\left( {{\beta _2}{\beta _1} + {\beta _3}{\beta _1}} \right)} \right)} \right\}\) On further simplification, \(z1 \times (z2 + z3)\) becomes, \({z_1} \times \left( {{z_2} + {z_3}} \right) = \left\{ {\left( {{\alpha _1}{\alpha _2} + i{\alpha _2}{\beta _1} + {\alpha _1}{\alpha _3} + i{\alpha _3}{\beta _1}} \right) + i\left( {{\beta _2}{\alpha _1} + {\beta _3}{\alpha _1}} \right) – \left( {{\beta _2}{\beta _1} + {\beta _3}{\beta _1}} \right)} \right\}\) \( = \left\{ {\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} – {\beta _2}{\beta _1} – {\beta _3}{\beta _1}} \right) + i\left( {{\beta _2}{\alpha _1} + {\alpha _2}{\beta _1} + {\beta _3}{\alpha _1} + {\alpha _3}{\beta _1}} \right)} \right\}\) \( = \left( {{\alpha _1}{\alpha _2} – {\beta _1}{\beta _2} + {\alpha _1}{\alpha _3} – {\beta _1}{\beta _3}} \right) + i\left( {{\alpha _3}{\beta _1} + {\alpha _2}{\beta _1} + {\alpha _1}{\beta _2} + {\alpha _1}{\beta _3}} \right)\) \( = \left( {{\alpha _1}{\alpha _2} – {\beta _1}{\beta _2}} \right) + i\left( {{\alpha _2}{\beta _1} + {\alpha _1}{\beta _2}} \right) + \left( {{\alpha _1}{\alpha _3} – {\beta _1}{\beta _3}} \right) + i\left( {{\alpha _3}{\beta _1} + {\alpha _1}{\beta _3}} \right)\) \( = \left( {{\alpha _1} + i{\beta _1}} \right)\left( {{\alpha _2} + i{\beta _2}} \right) + \left( {{\alpha _1} + i{\beta _1}} \right)\left( {{\alpha _3} + i{\beta _3}} \right)\) \( = \left( {{z_1} \times {z_2}} \right) + \left( {{z_1} \times {z_3}} \right)\) | Applicable |
| Multiplicative Inverse | \({\left( {{a_1} + i{\beta _1}} \right)^{ – 1}} = \frac{1}{{{a_1} + i{\beta _1}}}\) \( = \frac{1}{{{a_1} + i{\beta _1}}} \times \frac{{{a_1} – i{\beta _1}}}{{{a_1} – i{\beta _1}}}\) \( = \frac{{{a_1} – i{\beta _1}}}{{{{\left( {{a_1}} \right)}^2} – {{\left( {i{\beta _1}} \right)}^2}}}\) \( = \frac{{{a_1} – i{\beta _1}}}{{a_1^2 + \beta _1^2}}\) | Multiplicative inverse of \({a_1} + i{\beta _1}\) is \(\frac{{{a_1} – i{\beta _1}}}{{a_1^2 + \beta _1^2}}\) |
| Multiplicative Identity | \({z_1} \times A.Id. = {z_1}\) \(A.Id. = \frac{{{z_1}}}{{{z_1}}}\) \(A.Id. = 1\) | Multiplicative Identity of \({{z_1}}\) is \(1\) |
Mathematically, although the division of complex numbers is similar to real numbers, it is a lengthier process. This is because it is difficult to perform division using an imaginary number.
Method 1: Rationalising the Denominator
Step 1: Write the division of complex numbers as a fraction.
Step 2: Rationalise the denominator to remove the imaginary part of the divisor. This is done by multiplying the conjugate of the denominator with the numerator and the denominator.
Step 3: Simplify the expression using the algebraic identity \((a + b)(a – b) = {a^2} – {b^2}\).
Step 4: Substitute \({i^2} = – 1\).
Step 5: Simplify the numerator using the distributive property.
Step 6: Separate the real and imaginary parts in the resulting complex number.
Method 2: Using the Division Formula for Complex Numbers
The alternate method is to use the division of complex numbers formula directly.
Formula for the division of complex numbers \({z_1} = a + ib\) and \({z_2} = c + id\) is given by,
\(\frac{{{z_1}}}{{{z_2}}} = \frac{{a + ib}}{{c + id}}\)
\( = \frac{{(a + ib)(c – id)}}{{(c + id)(c – id)}}\)
\( = \frac{{(ac + bd) + i(bc – ad)}}{{{c^2} + {d^2}}}\)
\(\therefore \,\frac{{{z_1}}}{{{z_2}}} = \frac{{ac + bd}}{{{c^2} + {d^2}}} + i\left( {\frac{{bc – ad}}{{{c^2} + {d^2}}}} \right)\)
For any two complex numbers, \({z_1} = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\) and \({z_2} = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\), the formula for division in polar form is given by,
Q.1. Write \(z = \frac{{ – 1 + i3}}{{2 + i5}}\) in the form of \(a + ib\)
Ans:
\(\frac{{{z_1}}}{{{z_2}}} = \frac{{ac + bd}}{{{c^2} + {d^2}}} + i\left( {\frac{{bc – ad}}{{{c^2} + {d^2}}}} \right)\)
Here,
\({z_1} = – 1 + i3\)
\({z_2} = 2 + i5\)
\(\therefore \,\frac{{{z_1}}}{{{z_2}}} = \frac{{[( – 1) \times 2] + [3 \times 5]}}{{{2^2} + {5^2}}} + i\frac{{[3 \times 2] – [( – 1) \times 2]}}{{{2^2} + {5^2}}}\)
\( = \frac{{[ – 2] + 15}}{{4 + 25}} + i\frac{{6 – [ – 2]}}{{4 + 25}}\)
\(\therefore \,\frac{{ – 1 + i3}}{{2 + i5}} = \frac{{13}}{{29}} + i\frac{8}{{29}}\)
Q.2. Simplify: \((3 + i2)(4 + i5)\)
Ans:
\((3 – 2i)(4 + 5i) = 3 \times (4 + i5) + ( – i2)(4 + i5)\)
\( = 12 + i15 + ( – i8) – {i^2}10\)
\( = 12 + 7i + – 10 \times – 1\)
\( = 12 + 7i + 10\)
\( = 22 + 7i\)
Q.3. Simplify: \(\frac{{{{(1 – i)}^3}}}{{1 – {i^3}}}\)
Ans:
\(\frac{{{{(1 – i)}^3}}}{{1 – {i^3}}} = \frac{{{{(1 – i)}^3}}}{{(1 – i)\left( {1 + i + {i^2}} \right)}}\)
\( = \frac{{\left( {1 – {i^2}} \right)}}{i}\)
\( = \frac{{1 + {i^2} – i2}}{i}\)
\( = \frac{{ – i2}}{i}\)
\(\therefore \,\frac{{{{(1 – i)}^3}}}{{1 – {i^3}}} = – 2\)
Q.4. Expand and simplify: \((4 – i3)(2 + i3)(2 – i3)\)
Ans:
\((4 – i3)(2 + i3)(2 – i3) = (4 – i3)\left( {4 + i6 – i6 – {i^2}9} \right)\)
\( = (4 – i3)(4 – 9( – 1))\)
\( = (4 – i3)(4 + 9)\)
\( = (4 – i3) \times 13\)
\(\therefore \,(4 – i3)(2 + i3)(2 – i3) = 52 – i39\)
Q5. Subtract \((10 – i8)\) from \((13 – i7)\).
Ans:
\((13 – i7) – (10 – i8) = 13 – i7 – 10 + i8\)
\( = (13 – 10) – i7 + i8\)
\(\therefore \,(13 – i7) – (10 – i8) = 3 + i\)
Q6. If \(\frac{{{{(1 + i)}^2}}}{{2 – i}} = x + iy\), then find the value of \(x + iy\)
Ans:
\(x + iy = \frac{{{{(1 + i)}^2}}}{{2 – i}}\)
\( = \frac{{1 + 2i + {i^2}}}{{2 – i}}\)
\( = \frac{{2i}}{{2 – i}}\)
\( = \frac{{2i(2 + i)}}{{(2 – i)(2 + i)}}\)
\( = \frac{{4i + 2{i^2}}}{{4 – {i^2}}}\)
\( = \frac{{4i – 2}}{{4 + 1}}\)
\( = \frac{{ – 2}}{5} + \frac{{4i}}{5}\)
\( \Rightarrow x = \frac{{ – 2}}{5},\,y = \frac{4}{5}\)
\( \Rightarrow x + y = \frac{{ – 2}}{5} + \frac{4}{5}\)
\(\therefore \,x + y = \frac{2}{5}\)
Complex numbers can be represented in rectangular and polar forms. Similar to real numbers, arithmetic operations such as addition, subtraction, multiplication and division can also be performed on complex numbers.
Q.1. What are the mathematical operations of complex numbers?
Ans: The four operations that can be performed with complex numbers are addition, subtraction, multiplication, and division. However, subtraction is defined in terms of addition, and division in terms of multiplication.
Q.2. What is the addition of complex numbers?
Ans: While adding two or more complex numbers, the real parts and the imaginary numbers are added separately.
Q.3. Can you multiply real and imaginary numbers?
Ans: The short answer is yes. Mathematically, the multiplication happens only with the real number, but the imaginary unit \((i)\) is carried forward.
For example, \(5(12 – i3) = 5 \times 12 – 5 \times i3\)
\(\therefore \,5(12 – i3) = 60 – i15\)
Q.4. How do you solve complex numbers with multiplication?
Ans: While multiplying two complex numbers, remember to multiply each part of the first complex number with each part of the second complex number. Alternatively, remember the pneumonic ‘FOIL’ that expands to First, Outer, Inner, and Last that denotes the order in which pairs of numbers are multiplied.
Example: \((p + iq)(x + iy)\)
Here,
Q.5. Do complex numbers commute?
Ans: Complex numbers are commutative under addition and multiplication. This means that while adding or multiplying complex numbers, changing the order of the numbers does not alter the result.
Learn About Complex Numbers Here
We hope this detailed article on the Operations with Complex Numbers was helpful. If you have any doubts, let us know about them in the comment section below. Our team will get try to solve your queries at the earliest.
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