Answer
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Hint- First, we solve the question by separating powers of e and then by comparing the formula for modulus and argument of complex number with the given complex number. We will get our answer.
Complete step by step answer:
Given, complex number
$z = {\left( {{e^{3 - i\dfrac{\pi }{4}}}} \right)^3}$
Now, by separating powers
$ = {\left( {{e^3}.{e^{ - \dfrac{{i\pi }}{4}}}} \right)^3}$
So, multiplying by power
$ = {e^9}.{e^{ - 3\dfrac{{i\pi }}{4}}}$
As we know that, we can denote the complex number (z) in polar form
$z = \left| z \right|{e^{i\arg \left( z \right)}}$
${e^9}.{e^{ - 3\dfrac{{i\pi }}{4}}} = \left| z \right|{e^{i\arg \left( z \right)}}$
By comparing RHS and LHS we get
$\left| z \right| = {e^9}$ and
$\arg \left( z \right) = - \dfrac{{3\pi }}{4}$
Hence, the correct answer is D.
Note- Complex number is the number that can be expressed in the form of ‘a+bi’, where a and b are real numbers and i is iota i.e. $\sqrt { - 1} $. This is a direct question if you know the formula. So, remembering the formula is must and carefully moulding the question in the same form is very crucial for solving these types of questions.
Complete step by step answer:
Given, complex number
$z = {\left( {{e^{3 - i\dfrac{\pi }{4}}}} \right)^3}$
Now, by separating powers
$ = {\left( {{e^3}.{e^{ - \dfrac{{i\pi }}{4}}}} \right)^3}$
So, multiplying by power
$ = {e^9}.{e^{ - 3\dfrac{{i\pi }}{4}}}$
As we know that, we can denote the complex number (z) in polar form
$z = \left| z \right|{e^{i\arg \left( z \right)}}$
${e^9}.{e^{ - 3\dfrac{{i\pi }}{4}}} = \left| z \right|{e^{i\arg \left( z \right)}}$
By comparing RHS and LHS we get
$\left| z \right| = {e^9}$ and
$\arg \left( z \right) = - \dfrac{{3\pi }}{4}$
Hence, the correct answer is D.
Note- Complex number is the number that can be expressed in the form of ‘a+bi’, where a and b are real numbers and i is iota i.e. $\sqrt { - 1} $. This is a direct question if you know the formula. So, remembering the formula is must and carefully moulding the question in the same form is very crucial for solving these types of questions.
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