Question

How to do it and what's the answer?
`(4 - 2) - ( - 3 - i)`

Answer 1

`(4-2i)-(-3+i)`

When we are working with imaginary numbers, it is like working with an algebraic expression in which i will be the variable and can only added or substracted with expressions with the same variable i.

To solve this problem add all number without the i and also add all numbers with the imaginary portion.

`\begin{gathered} 4-2i-(-3+i) \\ 4-2i+3-i \\ 7-3i \end{gathered}`

When we talk about the product between imaginary numbers, we apply the same principal for algebraic however we mus remember the power derived from the i, here are some examples.

`\begin{gathered} i=\sqrt[]{-1} \\ i^2=(\sqrt[]{-1})^2=-1 \\ i^3=(\sqrt[]{-1})^3=-\sqrt[]{-1}=-i \\ i^4=(\sqrt[]{-1})^4=1 \end{gathered}`

remembering this, start by solving the product

`(2+i)(4-5i)`

distribute the factors

`\begin{gathered} 2\cdot(4-5i)+i(4-5i) \\ 8-10i+4i-5i^2 \\ 8-6i-5i^2 \end{gathered}`

apply the powers for i

`\begin{gathered} 8-6i-5\cdot(-1) \\ 8-6i+5 \\ 13-6i \end{gathered}`