Question

Write your answer as a complex number in standard form, I need help please

Answer 1

Answer:

10. $\frac{-3+4i}{25}$

11. $\frac{3+3i}{2}$

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.

$i^1 = i\\i^2 = -1\\i^3 = -i\\i^4 = 1$

Notice that $i^2$ and $i^4$ both give real number values of -1, and 1. Since we cannot combine real numbers and imaginary numbers with operations like $4-3i$ ≠ $i$, we will use exponents of imaginary numbers to convert and properties of algebra to make it possible.

For $\frac{i}{4-3i}$ we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose $\frac{4+3i}{4+3i} = 1$ and multiply.

$\frac{i}{4-3i} * \frac{4+3i}{4+3i} = \frac{i(4+3i)}{(4-3i)(4+3i} =\frac{4i+3i^2}{16-9i^2}$

Notice $i^2=-1$. We replace it and simplify the integers. Then write it in a+bi order.

$\frac{4i+3i^2}{16-9i^2}=\frac{4i+3(-1)}{16-9(-1)}= \frac{4i-3}{16+9}=\frac{-3+4i}{25}$.

We repeat the steps for the second problem.

For $\frac{-3+3i}{2i}$ we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose $\frac{2i}{2i} = 1$ and multiply.

$\frac{-3+3i}{2i}* \frac{2i}{2i} = \frac{(-3+3i)(2i)}{(2i)(2i)} =\frac{-6i+6i^2}{4i^2}$

Notice $i^2=-1$. We replace it and simplify the integers. Then write it in a+bi order.

$\frac{-6i+6i^2}{4i^2}=\frac{-6i+6(-1)}{4(-1)}=\frac{-6i-6}{-4}=\frac{-6-6i}{-4}= \frac{-2(3+3i)}{-2(2)}=\frac{3+3i}{2}$.

Answer 2

10. -0.12+0.16i


11. 1.5+1.5i