Answer:
10. [tex]\frac{-3+4i}{25}[/tex]
11. [tex]\frac{3+3i}{2}[/tex]
Step-by-step explanation:
A complex number is a number made of two parts: a real number and an imaginary number. We write them as a+bi.
- a is a real number
- bi is imaginary
We never use i as a variable in math because i is a symbol we use for the square root of -1. We need to remember how powers of i work because we will be multiplying them and exponents may be possible.
[tex]i^1 = i\\i^2 = -1\\i^3 = -i\\i^4 = 1[/tex]
Notice that [tex]i^2[/tex] and [tex]i^4[/tex] both give real number values of -1, and 1. Since we cannot combine real numbers and imaginary numbers with operations like [tex]4-3i[/tex] ≠ [tex]i[/tex], we will use exponents of imaginary numbers to convert and properties of algebra to make it possible.
For [tex]\frac{i}{4-3i}[/tex] we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose [tex]\frac{4+3i}{4+3i} = 1[/tex] and multiply.
[tex]\frac{i}{4-3i} * \frac{4+3i}{4+3i} = \frac{i(4+3i)}{(4-3i)(4+3i} =\frac{4i+3i^2}{16-9i^2}[/tex]
Notice [tex]i^2=-1[/tex]. We replace it and simplify the integers. Then write it in a+bi order.
[tex]\frac{4i+3i^2}{16-9i^2}=\frac{4i+3(-1)}{16-9(-1)}= \frac{4i-3}{16+9}=\frac{-3+4i}{25}[/tex].
We repeat the steps for the second problem.
For [tex]\frac{-3+3i}{2i}[/tex] we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose [tex]\frac{2i}{2i} = 1[/tex] and multiply.
[tex]\frac{-3+3i}{2i}* \frac{2i}{2i} = \frac{(-3+3i)(2i)}{(2i)(2i)} =\frac{-6i+6i^2}{4i^2}[/tex]
Notice [tex]i^2=-1[/tex]. We replace it and simplify the integers. Then write it in a+bi order.
[tex]\frac{-6i+6i^2}{4i^2}=\frac{-6i+6(-1)}{4(-1)}=\frac{-6i-6}{-4}=\frac{-6-6i}{-4}= \frac{-2(3+3i)}{-2(2)}=\frac{3+3i}{2}[/tex].
