
If $ z = 6{e^{i\dfrac{\pi }{3}}} $ then $ \left| {{e^{iz}}} \right| = $
A. $ \dfrac{1}{{{e^{\sqrt 2 }}}} $
B. $ \dfrac{1}{{{e^{2\sqrt 2 }}}} $
C. $ \dfrac{1}{{{e^{3\sqrt 3 }}}} $
D. $ \dfrac{1}{{{e^{1\sqrt 3 }}}} $
Answer
459.3k+ views
Hint: First of all we will expand exponential form and its power in terms of sine and cosine. Then will place the value for the given angle and then simplify for the required answer.
Complete step-by-step answer:
Take the given expression –
$ z = 6{e^{i\dfrac{\pi }{3}}} $
The above equation can be re-written as-
$ z = 6\left( {{e^{i\dfrac{\pi }{3}}}} \right) $
Expand the exponential form in sine and cosine function –
$ z = 6\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) $
Multiply with imaginary number “i” on both sides of the equation.
$ \Rightarrow iz = 6i\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) $
Multiply “i” inside the bracket on the left hand side of the equation.
$ \Rightarrow iz = 6\left( {i\cos \dfrac{\pi }{3} + {i^2}\sin \dfrac{\pi }{3}} \right) $
We know that - $ ({i^2}) = 1 $ , place in the above equation –
$ \Rightarrow iz = 6\left( {i\cos \dfrac{\pi }{3} - \sin \dfrac{\pi }{3}} \right) $
The above equation can be expressed as –
$ \Rightarrow iz = 6\left( { - \sin \dfrac{\pi }{3} + i\cos \dfrac{\pi }{3}} \right) $ .... (A)
Use the identity –
$ \left| {{e^z}} \right| = {e^{\operatorname{Re} z}} $
Similarly imaginary function can be defined as –
$ \left| {{e^{iz}}} \right| = {e^{\operatorname{Re} (iz)}} $
Place the value of equation (A) in the above equation –
$ \left| {{e^{iz}}} \right| = {e^{\operatorname{Re} \left[ {6\left( { - \sin \dfrac{\pi }{3} + i\cos \dfrac{\pi }{3}} \right)} \right] }} $
Real part in the above equation is
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - 6\left( {\sin \dfrac{\pi }{3}} \right)}}\]
Place the value of sine angle in the above equation –
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - 6\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}\]
Find the multiple in the above equation.
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - (3 \times 2)\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}\]
Common multiple from the numerator and denominator cancel each other.
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - (3\sqrt 3 )}}\]
Use the law of inverse exponent –
\[ \Rightarrow \left| {{e^{iz}}} \right| = \dfrac{1}{{{e^{(3\sqrt 3 )}}}}\]
So, the correct answer is “Option C”.
Note: Know the concepts of the complex numbers. It is the combination of real numbers and the imaginary numbers and since imaginary numbers are very difficult to understand and therefore they are complex numbers. Also, be good in multiples and simplifications of the equation. Remembering the square of the negative terms also gives the positive values.
The complex number consists of the real part and an imaginary part and is denoted by “Z”. It can be expressed as $ z = a + ib $ where “a” is the real part and “b” is the imaginary part. The modulus of the complex number is the length of the vector and can be expressed as $ r = \sqrt {{a^2} + {b^2}} $
Complete step-by-step answer:
Take the given expression –
$ z = 6{e^{i\dfrac{\pi }{3}}} $
The above equation can be re-written as-
$ z = 6\left( {{e^{i\dfrac{\pi }{3}}}} \right) $
Expand the exponential form in sine and cosine function –
$ z = 6\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) $
Multiply with imaginary number “i” on both sides of the equation.
$ \Rightarrow iz = 6i\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) $
Multiply “i” inside the bracket on the left hand side of the equation.
$ \Rightarrow iz = 6\left( {i\cos \dfrac{\pi }{3} + {i^2}\sin \dfrac{\pi }{3}} \right) $
We know that - $ ({i^2}) = 1 $ , place in the above equation –
$ \Rightarrow iz = 6\left( {i\cos \dfrac{\pi }{3} - \sin \dfrac{\pi }{3}} \right) $
The above equation can be expressed as –
$ \Rightarrow iz = 6\left( { - \sin \dfrac{\pi }{3} + i\cos \dfrac{\pi }{3}} \right) $ .... (A)
Use the identity –
$ \left| {{e^z}} \right| = {e^{\operatorname{Re} z}} $
Similarly imaginary function can be defined as –
$ \left| {{e^{iz}}} \right| = {e^{\operatorname{Re} (iz)}} $
Place the value of equation (A) in the above equation –
$ \left| {{e^{iz}}} \right| = {e^{\operatorname{Re} \left[ {6\left( { - \sin \dfrac{\pi }{3} + i\cos \dfrac{\pi }{3}} \right)} \right] }} $
Real part in the above equation is
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - 6\left( {\sin \dfrac{\pi }{3}} \right)}}\]
Place the value of sine angle in the above equation –
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - 6\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}\]
Find the multiple in the above equation.
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - (3 \times 2)\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}\]
Common multiple from the numerator and denominator cancel each other.
\[ \Rightarrow \left| {{e^{iz}}} \right| = {e^{ - (3\sqrt 3 )}}\]
Use the law of inverse exponent –
\[ \Rightarrow \left| {{e^{iz}}} \right| = \dfrac{1}{{{e^{(3\sqrt 3 )}}}}\]
So, the correct answer is “Option C”.
Note: Know the concepts of the complex numbers. It is the combination of real numbers and the imaginary numbers and since imaginary numbers are very difficult to understand and therefore they are complex numbers. Also, be good in multiples and simplifications of the equation. Remembering the square of the negative terms also gives the positive values.
The complex number consists of the real part and an imaginary part and is denoted by “Z”. It can be expressed as $ z = a + ib $ where “a” is the real part and “b” is the imaginary part. The modulus of the complex number is the length of the vector and can be expressed as $ r = \sqrt {{a^2} + {b^2}} $
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