Questions: Step 3: Simplify by writing the quotient in standard form a+bi. (2+3i)(7+2i)/(7-2i)(7+2i) = 14+(□)i+(□)i^2/(7-2i)(7+2i)

Transcript text: Step 3: Simplify by writing the quotient in standard form $\mathrm{a}+\mathrm{bi}$. \[ \frac{(2+3 i)(7+2 i)}{(7-2 i)(7+2 i)}=\frac{14+(\square) i+(\square) i^{2}}{(7-2 i)(7+2 i)} \]

Solution

Solution Steps

Step 1: Expand the numerator

Expand the numerator $ (2+3i)(7+2i) $: \[ (2+3i)(7+2i) = 2 \cdot 7 + 2 \cdot 2i + 3i \cdot 7 + 3i \cdot 2i \] \[ = 14 + 4i + 21i + 6i^2 \] \[ = 14 + 25i + 6i^2 \]

Step 2: Simplify $ i^2 $

Since $ i^2 = -1 $, substitute $ i^2 $ with $-1$: \[ 14 + 25i + 6(-1) = 14 + 25i - 6 \] \[ = 8 + 25i \]

Step 3: Expand the denominator

Expand the denominator $ (7-2i)(7+2i) $: \[ (7-2i)(7+2i) = 7 \cdot 7 + 7 \cdot 2i - 2i \cdot 7 - 2i \cdot 2i \] \[ = 49 + 14i - 14i - 4i^2 \] \[ = 49 - 4i^2 \]

Step 4: Simplify $ i^2 $ in the denominator

Substitute $ i^2 $ with $-1$: \[ 49 - 4(-1) = 49 + 4 \] \[ = 53 \]

Step 5: Write the quotient in standard form

Divide the simplified numerator by the denominator: \[ \frac{8 + 25i}{53} = \frac{8}{53} + \frac{25}{53}i \]