Complex numbers lend themselves readily to the basic mathematical
operations of addition, subtraction, multiplication, and division. A few
basic rules and definitions must be understood before considering these
operations.
Let us first examine the symbol $j$ associated with imaginary numbers. By definition,
$$\bbox[10px,border:1px solid grey]{ j = \sqrt{-1}} \tag{1}$$
thus
$$\bbox[10px,border:1px solid grey]{ j^2 = -1} \tag{2}$$
and
$$ j^4 = j^2 j^2 = (-1)(-1)=1$$
and so on. Further,
$${1 \over j} = ({1 \over j})({j \over j}) = ({j \over j^2})=({j \over -1}) = -j$$
and
$$\bbox[10px,border:1px solid grey]{{1 \over j}= -j}$$
Complex Conjugate
The conjugate or complex conjugate of a
complex number can be
found by simply changing the sign of the imaginary part in the
rectangular form or by using the negative of the angle of the polar form. For example, the conjugate of
is
as shown in
[Fig. 1]. The conjugate of
$$C = 2 \angle 30^\circ$$
is
$$C = 2 \angle -30^\circ$$
as shown in
[Fig. 2].
Fig. 1: Defining the complex conjugate of a complex number in rectangular form.
Fig. 2: Defining the complex conjugate of a complex number in polar form.
Reciprocal of complex numbers
The reciprocal of a complex number is 1 divided by the complex number. For example, the reciprocal of
is
and of $Z \angle \theta$,
$$ {1 \over Z \angle \theta}$$
We are now prepared to consider the four basic operations of
addition, subtraction, multiplication, and division with complex numbers.
Addition of complex numbers
To add two or more complex numbers, simply add the real and imaginary parts separately. For example, if
$C_1 = X_1 + j Y_1$ and $C_2 = X_2 + j Y_2$ then
$$\bbox[10px,border:1px solid grey]{C_1 + C_2 = X_1 + X_2 + j(Y_1+Y_2)} \tag{3}$$
There is really no need to memorize the equation. Simply set one above
the other and consider the real and imaginary parts separately, as shown
in Example 1.
Example 1:
a. Add $C_1 = 2 + j 4$ and $C_2 =3 +j 1$.
b. Add $C_1 = 3 +j 6$ and $C_2= -6 + j 3$.
Solution:
a. By Eq. (3),
$$C_1 +C_2= 2 + 3 + j (4 + 1) = 5+j5$$
Note
[Fig. 3],
b. By Eq. (3),
$$C_1 +C_2= 3 - 6 + j (6 + 3) = -3+j9$$
Note Fig. 4,
Fig. 3: Example 1(a)
Fig. 4: Example 1(b)
Subtraction of complex numbers
In subtraction, the real and imaginary parts are again considered separately. For example, if
$C_1 = X_1 + j Y_1$ and $C_2 = X_2 + j Y_2$ then
$$\bbox[10px,border:1px solid grey]{C_1 - C_2 = X_1 - X_2 + j(Y_1-Y_2)} \tag{4}$$
Again, there is no need to memorize the equation.
Example 2:
a. Subtract $C_2 = 1 + j 4$ from $C_1 =4 +j 6$.
b. Subtract $C_2 = -2 +j 5$ from $C_1= 3 + j 3$.
Solution:
a. By Eq. (4),
$$C_1 - C_2= 4 - 1 + j (6 - 4) = 3+j2$$
Note
[Fig. 5],
b. By Eq. (4),
$$C_1 +C_2= 3-(-2) + 6 + j (3 - 5) = 5-j2$$
Note
[Fig. 6],
Fig. 5: Example 2(a)
Fig. 6: Example 2(b)
Addition or subtraction cannot be performed in polar form unless the complex numbers have the same angle v or unless they differ only by multiples of $180^\circ$.
Example 3:
a. $2 \angle 45^\circ + 3 \angle 45^\circ= 5 \angle 45^\circ$
Note
[Fig. 7],
b. $2 \angle 0^\circ - 4 \angle 180^\circ = 6 \angle 0^\circ$
Note
[Fig. 8],
Fig. 7: Example 3(a)
Fig. 8: Example 3(b)
Multiplication of complex numbers
To multiply two complex numbers in rectangular form, multiply the
real and imaginary parts of one in turn by the real and imaginary parts
of the other. For example, if
$C_1 = X_1 + j Y_1$ and $C_2 = X_2 + j Y_2$ then
$C_1 \cdot C_2$:
$$X_1 + j Y_1 \\
\underline{ X_2 + j Y_2} \\
X_1X_2 + j Y_1X_2 \\
\underline{ + jY_2 X_1 + j^2 Y_2 Y_1}\\
X_1X_2 + j (Y_1X_2 + Y_2 X_1 ) - Y_2 Y_1$$
and
$$ \bbox[10px,border:1px solid grey]{C_1 \cdot C_2 = (X_1X_2 - Y_2 Y_1) + j (Y_1X_2 + Y_2 X_1 )} \tag{5}$$
Example 4:
a. Find $C_1 \cdot C_2$ if
$C_1 = 2 + j 3$ and $C_2= 5 + j 10$
b. Find $C_1 \cdot C_2$ if
$C_1 = -2 - j 3$ and $C_2= 4 - j 6$
Solution:
a. Using the format above of eq(5), we have
$$\begin{split}
C_1 \cdot C_2 &= [(2)(5)-(3)(10)]+j[(3)(5) + (10)(2)]\\
&= [10-30]+j[15 + 20]\\
&= -20+j35\\
\end{split}$$
b. Using the format above of eq(5), we have
$$\begin{split}
C_1 \cdot C_2 &= [(-2)(4)-(-3)(-6)]+j[(-2)(-6) + (4)(-3)]\\
&= [-8-18]+j[12 -12]\\
&= -26-j0\\
&= -26\\
&= -26 \angle {0^\circ} = 26\angle {180^\circ}\\
\end{split}$$
In polar form, the magnitudes are multiplied and the angles added
algebraically. For example, for
$C_1 = Z_1 \angle {\theta_1}$ and $C_2 = Z_2 \angle {\theta_2}$
we write
$C_1 \cdot C_2 = Z_1 Z_2 \angle {\theta_1 + \theta_2} $
Division of complex numbers
To divide two complex numbers in rectangular form, multiply the
numerator and denominator by the conjugate of the denominator and
the resulting real and imaginary parts collected. That is, if
$C_1 = X_1 + j Y_1$ and $C_2 = X_2 + j Y_2$ then
then
$$\begin{split}
{C_1 \over C_2} &= {X_1 + j Y_1 \over X_2 + j Y_2}\\
&= {X_1 + j Y_1 \over X_2 + j Y_2} {X_2 - j Y_2 \over X_2 - j Y_2}\\
&= {[X_1X_2 - Y_1Y_2]+j[X_1Y_1+X_2Y_1] \over X_2^2 + Y_2^2}\\
\end{split}$$
and
$$\bbox[10px,border:1px solid grey]{{C_1 \over C_2}={X_1X_2 + Y_1Y_2 \over X_2^2 + Y_2^2} +j{X_2Y_1-X_1Y_2 \over X_2^2 + Y_2^2}} \tag{6}$$
The equation does not have to be memorized if the steps above used to obtain it are employed. That is, first multiply the numerator by the complex conjugate of the denominator and separate the real and imaginary terms. Then divide each term by the sum of each term of the
denominator squared.
Example 1:
a. Find $C_1/C_2$ if $C_1= 1 + j 4$ and $C_2 = 4 + j 5$.
b. Find $C_1/C_2$ if $C_1= -4 - j 8$ and $C_2= 6 - j1$.
Solution:
a. By Eq. (6),
$$\begin{split}
{C_1 \over C_2} &= {(1)(4) + (4)(5) \over 4^2+5^2} + j{[(4)(4) - (1)(5)] \over 4^2+5^2}\\
&= {4 + 20 \over 16+25} + j{[16 - 5] \over 16+25}\\
&= {24 \over 41} + j{11 \over 41}\\
&= 0.585 + j0.268\\
\end{split}$$
b. By Eq. (6),
$$\begin{split}
{C_1 \over C_2} &= {(-4)(6) + (-8)(-1) \over 6^2+1^2} + j{[(6)(-8) - (-4)(-1)] \over 6^2+1^2}\\
&= {-24 + 8 \over 36+1} + j{[-48 - 4] \over 36+1}\\
&= {-16 \over 37} - j{52 \over 37}\\
&= -0.432 - j1.405\\
\end{split}$$
In polar form, division is accomplished by simply dividing the magnitude of the numerator by the magnitude of the denominator and subtracting the angle of the denominator from that of the numerator. That
is, for
$C_1 = Z_1 \angle {\theta_1}$ and $C_2 = Z_2 \angle {\theta_2}$
$${C_1 \over C_2} = {Z_1 \angle {\theta_1} \over Z_2 \angle {\theta_2}}$$
$${C_1 \over C_2} = {Z_1 \over Z_2} \angle {\theta_1 - \theta_2}$$
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