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Question: If \(z = \dfrac{{(7 - i)}}{{(3 - 4i)}}\)then \({z^{14}}\)is equal to \(A){2^7}\) \[B){(2i)^7}\] ...

If z=(7−i)(3−4i)z = \dfrac{{(7 - i)}}{{(3 - 4i)}}then z14{z^{14}}is equal to
A)27A){2^7}
B)(2i)7B){(2i)^7}
C)(2i)14C){(2i)^{14}}
D)(−2i)7D){( - 2i)^7}
E)−214E) - {2^{14}}

Explanation

Solution

First, complex numbers are the real and imaginary combined numbers as in the form of z=x+iyz = x + iy, where x and y are the real numbers and ii is the imaginary.
Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1
We ask us to solve the function z=(7−i)(3−4i)z = \dfrac{{(7 - i)}}{{(3 - 4i)}}and then find the value of zz so that we get easily the required z14{z^{14}}

Complete step by step answer:
Since from the given that we have z=(7−i)(3−4i)z = \dfrac{{(7 - i)}}{{(3 - 4i)}}and by the use of the conjugate complex values, now we need to multiply and divide with 3+4i3 + 4i(opposite sign imaginary function) respectively to the given numerator and denominator, thus we have z=(7−i)(3−4i)×3+4i3+4iz = \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=(7−i)(3−4i)×3+4i3+4i⇒21−3i+28i−4i29−12i+12i−16i2z = \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 i2=−1{i^2} = - 1as in the complex plane, substitute the value in the above equation we get z=21−3i+28i−4i29−12i+12i−16i2⇒21+25i+49+16z = \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=25+25i25⇒25(1+i)25⇒(1+i)z = \dfrac{{25 + 25i}}{{25}} \Rightarrow \dfrac{{25(1 + i)}}{{25}} \Rightarrow (1 + i)
Thus, we get the value of z=1+iz = 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 zab=(za)b{z^{ab}} = {({z^a})^b}
Now apply power 1414on both sides on z=1+iz = 1 + i then we get z14=(1+i)14{z^{14}} = {(1 + i)^{14}}
Giving the power terms inside the value with 14=2×714 = 2 \times 7then we get z14=[(1+i)2]7{z^{14}} = {[{(1 + i)^2}]^7}
Now applying the general formula (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2aband thus we get z14=[(1+i)2]7⇒z14=[1+2i+i]7⇒z14=[1+2i−1]7{z^{14}} = {[{(1 + i)^2}]^7} \Rightarrow {z^{14}} = {[1 + 2i + i]^7} \Rightarrow {z^{14}} = {[1 + 2i - 1]^7}
Further solving we get z14=[2i]7{z^{14}} = {[2i]^7}

So, the correct answer is “Option B”.

Note: The general power rule is zab=(za)b{z^{ab}} = {({z^a})^b}that it giving the product value inside the given function using the multiplication separation method.
Since Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1and 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.