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$i^{23} = i^{20+3}$ $= i^{20} \\times i^{3}$ $= (i^{4})^{5} \\times i^{…

Question

$i^{23} = i^{20+3}$
$= i^{20} \times i^{3}$
$= (i^{4})^{5} \times i^{3}$
$= 1^{5} \times i^{3}$
$= 1 \times i^{3}$
$= -i$
use the example as a model. simplify the expressions.
$i^{37} = i$
complete
$i^{52} = $
don
$i$
$-i$
$1$
$-1$

Explanation:

Step1: Split exponent into sum

$i^{82}=i^{80+2}$

Step2: Rewrite as product of powers

$=i^{80} \times i^{2}$

Step3: Express as power of $i^4$

$=(i^4)^{20} \times i^{2}$

Step4: Substitute $i^4=1$

$=1^{20} \times i^{2}$

Step5: Simplify $1^{20}$ and $i^2$

$=1 \times (-1)$

Step6: Final multiplication

$=-1$

Answer:

$-1$