Question

# If $z=\frac{7-i}{3-4i}$, then ${z}^{14}$ is equal to

A

${2}^{7}$

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B

${2}^{7}i$

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C

${2}^{14}i$

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D

$-{2}^{7}i$

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E

$-{2}^{14}$

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Solution

## The correct option is D $-{2}^{7}i$Explanation for the correct option.Step 1. Simplify the complex number $z$.In $z=\frac{7-i}{3-4i}$, multiply the numerator and denominator by $3+4i$.$\begin{array}{rcl}z& =& \frac{7-i}{3-4i}·\frac{3+4i}{3+4i}\\ & =& \frac{\left(7-i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\\ & =& \frac{21+28i-3i-4{i}^{2}}{9-16{i}^{2}}\\ & =& \frac{21+25i-4\left(-1\right)}{9-16\left(-1\right)}\left[{i}^{2}=-1\right]\\ & =& \frac{21+25i+4}{9+16}\\ & =& \frac{25+25i}{25}\\ & =& 1+i\end{array}$Step 2. Find the value of ${Z}^{2}$So, $Z=1+i$. Now ${Z}^{2}$ is given as:$\begin{array}{rcl}{Z}^{2}& =& {\left(1+i\right)}^{2}=1+2i+{i}^{2}\\ & =& 1+2i+\left(-1\right)\left[{i}^{2}=-1\right]\\ & =& 1+2i-1\\ & =& 2i\end{array}$Step 3. Find the value of ${Z}^{14}$.Now, ${Z}^{14}$ is given as:$\begin{array}{rcl}{Z}^{14}& =& {\left({Z}^{2}\right)}^{7}\\ & =& {\left(2i\right)}^{7}\\ & =& {2}^{7}{i}^{7}\\ & =& {2}^{7}{i}^{4+3}\\ & =& {2}^{7}{i}^{4}{i}^{3}\left[{a}^{m+n}={a}^{m}{a}^{n}\right]\\ & =& {2}^{7}×1×\left(-i\right)\left[{i}^{4}=1,{i}^{3}=-i\right]\\ & =& -{2}^{7}i\end{array}$So the value of ${z}^{14}$ is $-{2}^{7}i$.Hence, the correct option is D.

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