Question

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

• A. $2^7$
• B. $2^{7}i$
• C. $2^{14}i$
• D. $-2^{7}i$
• E. $-2^{14}$

Answer 1

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 $3+4i$.

$\begin{array}{rcl}z& =& \frac{7-i}{3-4i}·\frac{3+4i}{3+4i}\\ & =& \frac{\left(7-i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\\ & =& \frac{21+28i-3i-4i^2}{9-16i^2}\\ & =& \frac{21+25i-4(-1)}{9-16(-1)}\left[i^2=-1\right]\\ & =& \frac{21+25i+4}{9+16}\\ & =& \frac{25+25i}{25}\\ & =& 1+i\end{array}$

Step 2. Find the value of $Z^2$

So, $Z=1+i$. Now $Z^2$ is given as:

$\begin{array}{rcl}{Z}^2& =& {(1+i)}^2=1+2i+i^2\\ & =& 1+2i+(-1)\left[i^2=-1\right]\\ & =& 1+2i-1\\ & =& 2i\end{array}$

Step 3. Find the value of $Z^{14}$.

Now, $Z^{14}$ is given as:

$\begin{array}{rcl}{Z}^{14}& =& {\left(Z^2\right)}^7\\ & =& {\left(2i\right)}^7\\ & =& 2^7{i}^7\\ & =& 2^7{i}^{4+3}\\ & =& 2^7{i}^4{i}^3\left[a^{m+n}=a^m{a}^n\right]\\ & =& 2^7\times 1\times \left(-i\right)\left[i^4=1,i^3=-i\right]\\ & =& -2^{7}i\end{array}$

So the value of $z^{14}$ is $-2^{7}i$.

Hence, the correct option is D.