MULTIPLICATION OF COMPLEX NUMBERS

Two numbers can be multiplied using this illustration:
$(a+bi)(c+di)=a(c+di)+bi(c+di)$
$=ac+adi+bci+bdi^2$

Example 1
Solve $(8+6i)(4-10i)$

Solution:
Now, let expand the brackets
$8(4-10i)+6i(4-10i)$
$32-80i+24i-60i^2$

$32-80i+24i-60(-1)$
$32-80i+24i+60$

To add complex numbers, we add real and imaginary numbers separately
$(32+60)+(-80+24)i$
$92-56i$

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

Solution:
$2(3-i)+i(3-i)$
$6-2i+3i-i^2$

If $i^2=-1$, then
$6-2i+3i-(-1)$
$6+i+1$
$7+i$

Example 3
Solve $(2+i)(2-i)$ 

Solution:
This is kind of difference of two square .
$2(2-i)+i(2-i)$
$4-2i+2i-i^2$
$4-(-1)$
$4+1=5$

Example 4
Solve $(6+9i)(4-6i)$

Solution:
$6(4-6i)+9i(4-6i)$
$24-36i+36i-54i^2$
$24-54(-1)$
$24+54=78$

Example 5
Given that $Z_1=2+i$, $Z_2=1-2i$ and $Z_3=1+i$, find
i) $Z_1Z_2$
ii) $Z_2Z_3$
iii) $Z_1Z_3$
iv) $Z_1Z_2Z_3$

Solution:
i) By substitution
$(2+i)(1-2i)$

Now, let's expand the brackets 
$2(1-2i)+i(1-2i)$
$2-4i+i-2i^2$
$2-3i-2(-1)$
$2-3i+2$
$4-3i$

ii) Now, let multiply $Z_2Z_3$ 

By substitution
$(1-2i)(1+i)$
$1(1+i)-2i(1+i)$
$1+i-2i-2i^2$
$1-i-2(-1)$
$1-i+2$
$3-i$

iii) $(2+i)(1+i)$
$2(1+i)+i(1+i)$
$2+2i+1i+i^2$
$2+3i+(-1)$
$2+3i-1$
$1+3i$

iv) We know from (i) that $Z_1Z_2$ is $4-3i$. Now let multiply $Z_3$ to $4-3i$

$(4-3i)(1+i)$
$4(1+i)-3i(1+i)$
$4+4i-3i-3i^2$
$4+i-3(-1)$
$4+i+3$
$7+i$

That is all for now. In the next post, we will continue our series on complex nembers when we look at division of complex numbers.

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