Question

If \[z = r{e^{i\theta }}\] , then what is the value of \[\left| {{e^{iz}}} \right|\]?
A. \[{e^{r sin\theta }}\]
B. \[{e^{ - r sin\theta }}\]
C. \[{e^{ - r cos\theta }}\]
D. \[{e^{r cos\theta }}\]

Answer 1

Hint: First, convert the complex number into the polar form. Then multiply the polar form of complex number by the imaginary number \[i\] and simplify the complex number. In the end, take the modulus of the complex number to get the required answer.
Formula used :
The polar form of a complex number \[z = a + ib\] is \[z = r\left( {cos\theta + isin\theta } \right)\], where \[r = \sqrt {{a^2} + {b^2}} \].
\[{e^{i\theta }} = \left( {cos\theta + isin\theta } \right)\]
\[\left| {{e^{ix}}} \right| = 1\] for all \[x \in R\].
\[{a^{\left( {m + n} \right)}} = {a^m}{a^n}\]
Complete step by step solution:
The given exponential form of a complex number is \[z = r{e^{i\theta }}\].
Let’s convert the exponential form into the polar form of a complex number.
\[r{e^{i\theta }} = r\left( {cos \theta + i sin \theta } \right)\]
\[ \Rightarrow \]\[z = r\left( {cos \theta + i sin \theta } \right)\]

Now multiply both sides of the above equation by \[i\].
 \[iz = ir\left( {cos \theta + i sin \theta } \right)\]
\[ \Rightarrow \]\[iz = r\left( {icos \theta + {i^2} sin \theta } \right)\]
\[ \Rightarrow \]\[iz = r\left( {icos \theta - sin \theta } \right)\] [ Since \[{i^2} = - 1\]]
\[ \Rightarrow \]\[iz = - rsin \theta + ircos \theta \]
Now take the exponential on both sides.
\[{e^{iz}} = {e^{ - rsin \theta + ircos \theta }}\]
Apply the exponent property \[{a^{\left( {m + n} \right)}} = {a^m}{a^n}\].
\[{e^{iz}} = {e^{ - rsin \theta }}{e^{ircos \theta }}\]

Take the modulus on both sides of the above equation.
\[\left| {{e^{iz}}} \right| = \left| {{e^{ - rsin \theta }} \times {e^{ircos \theta }}} \right|\]
\[ \Rightarrow \]\[\left| {{e^{iz}}} \right| = \left| {{e^{ - rsin \theta }}} \right| \times \left| {{e^{ircos \theta }}} \right|\]
\[ \Rightarrow \]\[\left| {{e^{iz}}} \right| = {e^{ - rsin \theta }} \times \left| {{e^{ircos \theta }}} \right|\]
Now apply the property \[\left| {{e^{ix}}} \right| = 1\] for all \[x \in R\] on right-hand side.
\[\left| {{e^{iz}}} \right| = {e^{ - rsin \theta }} \times 1\]
\[ \Rightarrow \]\[\left| {{e^{iz}}} \right| = {e^{ - rsin \theta }}\]

Hence the correct option is B.
Note: The different types of a complex number are:
Rectangular form: \[z = a + ib\]
Polar form: \[z = r\left( {cos\theta + isin\theta } \right)\], where \[r = \sqrt {{a^2} + {b^2}} \]
Exponential form: \[z = r{e^{i\theta }}\]