Question

1. simplify the expression. write your answer in terms of i.\\(\sqrt{-53}\\)\
2. simplify the term.\\(i^{179}\\)\
3. simplify the term.\\(i^{132}\\)\
4. simplify the term.\\(i^{110}\\)\
5. simplify the expression:\\(12 + \sqrt{-4}\\)\
6. simplify the expression:\\(3 - \sqrt{-36}\\)

Explanation:

Problem 3: Simplify $\boldsymbol{\sqrt{-53}}$

Step1: Rewrite using $i$

Recall that $\sqrt{-a} = \sqrt{a} \cdot i$ for $a > 0$.
$\sqrt{-53} = \sqrt{53 \cdot (-1)} = \sqrt{53} \cdot \sqrt{-1}$

Step2: Substitute $\sqrt{-1} = i$

$\sqrt{53} \cdot i = i\sqrt{53}$

Step1: Find the cycle of $i$

The powers of $i$ repeat every 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, then cycle.

Step2: Divide exponent by 4

Divide 179 by 4: $179 \div 4 = 44$ with remainder $3$ (since $44 \times 4 = 176$, $179 - 176 = 3$).

Step3: Simplify using the cycle

$i^{179} = i^{4 \times 44 + 3} = (i^4)^{44} \cdot i^3$
Since $i^4 = 1$, this becomes $1^{44} \cdot i^3 = 1 \cdot (-i) = -i$ (because $i^3 = -i$).

Step1: Use the cycle of $i$

Powers of $i$ cycle every 4: $i^4 = 1$.

Step2: Divide exponent by 4

$132 \div 4 = 33$ with remainder $0$.

Step3: Simplify

$i^{132} = i^{4 \times 33} = (i^4)^{33} = 1^{33} = 1$ (since $i^4 = 1$).

Answer:

$i\sqrt{53}$

Problem 4: Simplify $\boldsymbol{i^{179}}$