Question

If $z = \dfrac{{(7 - i)}}{{(3 - 4i)}}$then ${z^{14}}$is equal to
$A){2^7}$
\[B){(2i)^7}\]
$C){(2i)^{14}}$
$D){( - 2i)^7}$
$E) - {2^{14}}$

Answer 1

Hint: First, complex numbers are the real and imaginary combined numbers as in the form of $z = x + iy$, where x and y are the real numbers and $i$ is the imaginary.
Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$
We ask us to solve the function $z = \dfrac{{(7 - i)}}{{(3 - 4i)}}$and then find the value of $z$ so that we get easily the required ${z^{14}}$

Complete step by step answer:
Since from the given that we have $z = \dfrac{{(7 - i)}}{{(3 - 4i)}}$and by the use of the conjugate complex values, now we need to multiply and divide with $3 + 4i$(opposite sign imaginary function) respectively to the given numerator and denominator, thus we have $z = \dfrac{{(7 - i)}}{{(3 - 4i)}} \times \dfrac{{3 + 4i}}{{3 + 4i}}$(canceling both terms will get the original value)
Further solving with the multiplication operation, we have, $z = \dfrac{{(7 - i)}}{{(3 - 4i)}} \times \dfrac{{3 + 4i}}{{3 + 4i}} \Rightarrow \dfrac{{21 - 3i + 28i - 4{i^2}}}{{9 - 12i + 12i - 16{i^2}}}$
But since we know that ${i^2} = - 1$as in the complex plane, substitute the value in the above equation we get $z = \dfrac{{21 - 3i + 28i - 4{i^2}}}{{9 - 12i + 12i - 16{i^2}}} \Rightarrow \dfrac{{21 + 25i + 4}}{{9 + 16}}$
Further solving we get, $z = \dfrac{{25 + 25i}}{{25}} \Rightarrow \dfrac{{25(1 + i)}}{{25}} \Rightarrow (1 + i)$
Thus, we get the value of $z = 1 + i$
Now we need to apply the power value of fourteen to get the result, but before we need to know about the power rule of the product, which is ${z^{ab}} = {({z^a})^b}$
Now apply power $14$on both sides on $z = 1 + i$ then we get ${z^{14}} = {(1 + i)^{14}}$
Giving the power terms inside the value with $14 = 2 \times 7$then we get ${z^{14}} = {[{(1 + i)^2}]^7}$
Now applying the general formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$and thus we get ${z^{14}} = {[{(1 + i)^2}]^7} \Rightarrow {z^{14}} = {[1 + 2i + i]^7} \Rightarrow {z^{14}} = {[1 + 2i - 1]^7}$
Further solving we get ${z^{14}} = {[2i]^7}$

So, the correct answer is “Option B”.

Note: The general power rule is ${z^{ab}} = {({z^a})^b}$that it giving the product value inside the given function using the multiplication separation method.
Since Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$and without this value, we cannot solve the given problem also.
In conjugation, we only need to multiply and divide with opposite sign values in imaginary.