$\frac{1}{5 + 2 ni} \cdot \frac{1}{5 + 2 ni} = \frac{1}{4 i ^{2} n ^{2} + 20 in + 25}$

Explanation

\frac{1}{5 + 2 ni} \cdot \frac{1}{5 + 2 ni} \overset{◯}{1} \frac{1}{4 i ^{2} n ^{2} + 20 in + 25}

①

Multiply $\frac{1}{5 + 2 in}$ by $\frac{1}{5 + 2 in}$ to get $\frac{1}{4 i ^{2} n ^{2} + 20 in + 25}$ .

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

\frac{1}{5 + 2 in} \cdot \frac{1}{5 + 2 in} Step 1 \frac{1 \cdot 1}{( 5 + 2 in ) \cdot ( 5 + 2 in )} = Step 2 \frac{1}{25 + 10 in + 10 in + 4 i ^{2} n ^{2}} = \frac{1}{4 i ^{2} n ^{2} + 20 in + 25}

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5+2ni1​⋅5+2ni1​=4i2n2+20in+251​

$\frac{1}{5 + 2 ni} \cdot \frac{1}{5 + 2 ni} = \frac{1}{4 i ^{2} n ^{2} + 20 in + 25}$