Question

How do you simplify $\left( {6 + 2i} \right){{ }}\left( {5 - 3i} \right)?$

Answer 1

In this given question we have given two pairs of complex coordinates for which we need to find an equation of multiplication of both the points which will be in simplified form. You can use the distributive (also known as FOIL) law to multiply the two brackets together remembering the property that ${i^2} = - 1$; Finally we get the required answer.

Complete step by step answer: It can be performed in these following steps by attempting one by one as to be followed:
Step 1: Distribute (or FOIL) to remove the parentheses.
Step 2: Simplify the powers of i, specifically remember that ${i^2} = - 1$.
Step 3: Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers.
Thus we will perform our analysis for the following example as shown below:
Our Example has been stated as such:
Multiply: $\left( {6 + 2i} \right)\left( {5 - 3i} \right)$
Following step by step procedure we drive at the conclusion as proceeded further:
Step 1: Distributing to remove the parentheses of our given problem statement that is shown below:
$\Rightarrow (6 \times 5) - (3 \times 6i) + (2 \times 5i) - (3i \times 2i)$
Step 2: Simplifying the powers of $i$, specifically remembering that ${i^2} = - 1$ we will arrive at simplification as follows:
$\Rightarrow 30 - 18i + 10i + 6$
Step 3: Combining the like terms, that is combining the real numbers with real numbers and the imaginary numbers with imaginary numbers. We can derive the final result.
$\Rightarrow 36 - 8i$
Thus the most simplified form of our equation is being derived as $36 - 8i$.

Note:
We can also find the product of the two complex numbers by coordinate method using graphs or by De Moivre's theorem and the geometric visualization method. Some of them have been illustrated below:
De Moivre's theorem
If you're using complex numbers, then every polynomial equation of degree k yields exactly k solution. So, when we're expecting to find multiplication of complex numbers. De Moivre's theorem uses the fact that we can write any complex number as:
${{\rho }}{{{e}}^{{{i\theta }}}}{{ = \rho (cos(\theta ) + i sin(\theta ))}}$ , and it states that, if
${{z = \rho (cos(\theta ) + i sin(\theta ))}}$ , then
${{{z}}^{{n}}}{{ = }}{{{\rho }}^{{n}}}{{(cos(n\theta ) + i sin(n\theta ))}}$
Geometric visualization method
Let z1 and z2 be two complex numbers.
By rewriting in exponential form,

{z_1} = {r_1}{e^{i{\theta _1}}} \\\ {z_2} = {r_2}{e^{i{\theta _2}}} \\\ \right\\}$$ So, $$ \Rightarrow {z_1}{z_2} = {r_1}{e^{i{\theta _1}}}{r_2}{e^{i{\theta _2}}}$$ $$ \Rightarrow ({r_1}{r_2}){e^{i({\theta _1} + {\theta _2})}}$$ Hence, the product of two complex numbers can be geometrically interpreted as the combination of the product of their absolute values $$({r_1}{r_2})$$ and the sum of their angles $$({\theta _1} + {\theta _2})$$ as shown above.