Question

How do you simplify \[6i\left( 3+2i \right)\]?

Answer 1

Hint: We have been given a multiplication of complex numbers. We first try to find the sign of the multiplication from the coefficients. Then we multiply the constants and the complex digits separately. There will be two multiplications in total. We multiply them to find the solution of \[6i\left( 3+2i \right)\]. We also use the identity value of complex numbers where ${{i}^{2}}=-1$.

Complete step by step solution:
We have been given a multiplication of two terms \[6i\left( 3+2i \right)\].
The terms being $6i$ and \[\left( 3+2i \right)\]. Both of the terms are positive which makes the whole multiplication positive.
Therefore, the sign is known to us. We just need to find the multiplication of $6i$ and \[\left( 3+2i \right)\].
We break the bracket and find two multiplications.
So, \[6i\left( 3+2i \right)=\left( 6i \right)\left( 3 \right)+\left( 6i \right)\left( 2i \right)\].
For the first multiplication we get \[\left( 6i \right)\left( 3 \right)=18i\].
For the second multiplication the terms are constant and complex digits all being in multiplication.
Therefore, $6i\times 2i=6\times i\times 2\times i$.
We multiply the constants first and get $6\times 2=12$.
Then we multiply the variables and get $i\times i={{i}^{2}}$.
The total multiplication becomes $6i\times 2i=6\times i\times 2\times i=12{{i}^{2}}$.
We know that the value ${{i}^{2}}=-1$. Putting the value, we get $12{{i}^{2}}=-12$
The final form of the total multiplication is \[6i\left( 3+2i \right)=\left( 6i \right)\left( 3 \right)+\left( 6i \right)\left( 2i \right)=18i-12\].
The simplified form of \[6i\left( 3+2i \right)\] is \[18i-12\].

Note: For any multiplication and to find the appropriate signs for that we can follow the rule where multiplication of same signs gives positive result and multiplication of opposite signs give negative result. We can also multiply all the constants and the variables together to make the process simple.