ADDITION AND SUBTRACTION OF COMPLEX NUMBERS

Complex number, as I told you before, is the sum of real and imaginary numbers. And, as is with real numbers, we can also add or subtract complex numbers.

Two complex numbers are added by adding separately the two real parts and the two imaginary parts.

For example
$(a+bi)+(c+di)=(a+c)+(b+d)i$

In the same token, two complex numbers are substrates by subtracting separately the two real parts and the two imaginary parts.

For example
$(a+bi)-(c+di)=(a-c)+(b-d)i$

Using this fact, let solve some examples.

Example 1
Solve $(3-4i)+(-2+6i)$

Solution:
Now, let add real number separately and imaginary numbers separately
$(3-2)+(-4i+6i)$
$1+2i$

Example 2
Solve $(4-2i)+(3+7i)$

Solution:
$(4+3)+(7i-2i)$
$7+5i$

Example 3
Solve $(10+8i)-(6-4i)$

Solution:
Unlike the previous examples, we first expand the brackets
$10+8i-6+4i$
$10-6+8i+4i$
$4+12i$

Example 4
Solve the complex numbers $(2-4i)-3(5-3i)$

Solution:
First, we expand the brackets
$2-4i-15+9i$
$2-15-4i+9i$
$-13+5i$

Now, move to a more interesting example

Example 5
Given that $Z_1=4+8i$ and $Z_2=6-2i$, Find
i) $Z_1+Z_2$
ii) $Z_1–Z_2$
iii) $Z_2-Z_1$.

Solution:
i) By substitution
$(4+8i)+(6-2i)$
$(4+6)+(8i-2i)$
$10+6i$

ii) Again, by substitution
$(4+8i)-(6-2i)$

By expanding the brackets
$4+8i-6+2i$
$4-6+8i+2i$
$-2+10i$.

iii) By substitution
$(6-2i)-(4+8i)$

Expanding the brackets, we have
$6-2i-4-8i$
$6-4-2i-8i$
$2-10i$

There you have it. In our next post, we will be multiplying complex numbers. Meanwhile, if you have got question relating to complex numbers, do well to ask our telegram community
Help us grow our readership by sharing this post