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Question: Let \( f\left( z \right) = \sin z \) and \( g\left( z \right) = \cos z \) . If \( * \) denotes a com...

Let f(z)=sin⁡zf\left( z \right) = \sin z and g(z)=cos⁡zg\left( z \right) = \cos z . If ∗* denotes a composition of the function then show that (f+ig)∗(f−ig)=ie−eiz\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} .
i=−1i = \sqrt { - 1}

Explanation

Solution

Hint : Since we have to show (f+ig)∗(f−ig)=ie−eiz\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} and for this we will write (f+ig)∗(f−ig)\left( {f + ig} \right) * \left( {f - ig} \right) in the composition form and it will be F(G(z))F\left( {G\left( z \right)} \right) . From here F(z)=f+igF\left( z \right) = f + ig and G(z)=f−igG\left( z \right) = f - ig . Then we will solve both the composite equation and then in the end putting the values we will get the solution.
Formula used:
The algebraic formula is given by,
cos⁡z−isin⁡z=e−iz\cos z - i\sin z = {e^{ - iz}}
i=−1i = \sqrt { - 1}
Here, ii will be the iota.

Complete step-by-step answer :
Here, in this question, we have the values of the function given as f(z)=sin⁡zf\left( z \right) = \sin z and g(z)=cos⁡zg\left( z \right) = \cos z .
And we have to prove (f+ig)∗(f−ig)=ie−eiz\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}
Now taking the LHS of the upper function, we get
⇒(f+ig)∗(f−ig)\Rightarrow \left( {f + ig} \right) * \left( {f - ig} \right)
And it can also be written in the composition form, so it will be
⇒F(z)∗G(z)\Rightarrow F\left( z \right) * G\left( z \right)
And it will be equal to
⇒F(G(z))\Rightarrow F\left( {G\left( z \right)} \right)
Since, F(z)=f+igF\left( z \right) = f + ig
So, on substituting the values, we get
⇒F(z)=(sin⁡z+icos⁡z)\Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right)
Multiplying and dividing the above equation with the same function, ii
We get,
⇒F(z)=(sin⁡z+icos⁡z)ii\Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right)\dfrac{i}{i}
And on solving it, we get
⇒F(z)=isin⁡z+i2cos⁡zi\Rightarrow F\left( z \right) = \dfrac{{i\sin z + {i^2}\cos z}}{i}
Since we know i=−1i = \sqrt { - 1} , hence
⇒F(z)=isin⁡z−cos⁡zi\Rightarrow F\left( z \right) = \dfrac{{i\sin z - \cos z}}{i}
Taking the negative sign common, we get
⇒F(z)=−1i(cos⁡z−isin⁡z)\Rightarrow F\left( z \right) = - \dfrac{1}{i}\left( {\cos z - i\sin z} \right)
Also, from the formula, we know that cos⁡z−isin⁡z=e−iz\cos z - i\sin z = {e^{ - iz}} . So by using it we get
⇒F(z)=−1ie−iz\Rightarrow F\left( z \right) = - \dfrac{1}{i}{e^{ - iz}}
Again multiplying and dividing the above equation with the same function, ii
We get
⇒F(z)=−1i⋅iie−iz\Rightarrow F\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{ - iz}}
And on solving it we get
⇒F(z)=−1⋅ii2e−iz\Rightarrow F\left( z \right) = - \dfrac{{1 \cdot i}}{{{i^2}}}{e^{ - iz}}
Hence, it will be equal to
⇒F(z)=ie−iz\Rightarrow F\left( z \right) = i{e^{ - iz}}
Now we will solve, G(z)=f−igG\left( z \right) = f - ig
So, on substituting the values, we get
⇒G(z)=(sin⁡z−icos⁡z)\Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right)
Multiplying and dividing the above equation with the same function, ii
We get,
⇒G(z)=(sin⁡z−icos⁡z)ii\Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right)\dfrac{i}{i}
And on solving it, we get
⇒G(z)=isin⁡z−i2cos⁡zi\Rightarrow G\left( z \right) = \dfrac{{i\sin z - {i^2}\cos z}}{i}
Since we know i=−1i = \sqrt { - 1} , hence
⇒G(z)=isin⁡z+cos⁡zi\Rightarrow G\left( z \right) = \dfrac{{i\sin z + \cos z}}{i}
Also, from the formula, we know that cos⁡z−isin⁡z=e−iz\cos z - i\sin z = {e^{ - iz}} . So by using it we get
⇒G(z)=eizi\Rightarrow G\left( z \right) = \dfrac{{{e^{iz}}}}{i}
Again multiplying and dividing the above equation with the same function, ii
We get
⇒G(z)=−1i⋅iieiz\Rightarrow G\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{iz}}
And on solving it we get
⇒G(z)=−1⋅i−1eiz\Rightarrow G\left( z \right) = - \dfrac{{1 \cdot i}}{{ - 1}}{e^{iz}}
Hence, it will be equal to
⇒G(z)=−ieiz\Rightarrow G\left( z \right) = - i{e^{iz}}
So from the upper part of the solution,
⇒F(G(z))=iei⋅(G(z))\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( {G\left( z \right)} \right)}}
And on substituting the values, we get
⇒F(G(z))=iei⋅(−ieiz)\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( { - i{e^{iz}}} \right)}}
By using identities, we get the equation as
⇒F(G(z))=ie−i2eiz\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{ - {i^2}{e^{iz}}}}
And it can also be written as
⇒F(G(z))=ieeiz\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{{e^{iz}}}}
Since, we can see that LHS is equal to RHS,
Therefore, it is proved that (f+ig)∗(f−ig)=ie−eiz\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} will be equal.
So, the correct answer is “ (f+ig)∗(f−ig)=ie−eiz\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} ”.

Note : Solving composite functions means we have to find the composition of two functions. By rewriting the composite function in a different form we will be able to reach close to the answer. And lastly, by substituting the values we get it.