So if I have a function like the following, where $f$ is a complex function and B is real: $$S(\lambda) = 1 + iBf(\lambda)$$ Suppose that I know that its modulus must be equal to 1, therefore:

Now how do I determine the conditions for the real part and its imaginary part? I know that: $$|1 + iBf(\lambda)|^2 = 1$$

From here: $$|1 + iB [Re(f(\lambda)) + iIm(f(\lambda)]|^2 = 1$$

Then I can say that the imaginary and real parts of the function would be: $$Im(S(\lambda)) = 1+ BRe(f(\lambda))$$ $$Re(S(\lambda)) = -BIm(f(\lambda))$$

Is my reasoning correct?

  • 1
    $\begingroup$ Don't forget the $1$ on the real part! $\endgroup$ – cmk May 24 '19 at 17:22

For the modulus to be $1,$ you only need the product $Bf$ to be $\pm1,$ for since $$|1+iBf|=1,$$ it follows that $$(1+iBf)(1-iBf)=1+(Bf)^2=1,$$ and the result I claimed follows.

Of course I assumed $B$ and $f$ are real, since you do not say anything about them. Otherwise what I say above needs to be modified to be true.

OK, you have specified that $f(\lambda)$ is complex valued. Thus, if you write $f=a+ib,$ where $a=a(\lambda),\,\, b=b(\lambda).$ Then we have that $$1+iBf=1+iB(a+ib)=1-Bb+iBa.$$

For its modulus to be $1,$ we must have $$(1-Bb+iBa)(1-Bb-iBa)=1,$$ or $$(1-Bb)^2+(Ba)^2=1.$$ This gives $$B=\frac{2\Im f(\lambda)}{|f(\lambda)|^2}.$$

  • $\begingroup$ I meant $f$ to be complex and B to be real $\endgroup$ – daljit97 May 24 '19 at 19:25
  • $\begingroup$ @daljit97 Is $\lambda$ real or complex too? $\endgroup$ – Allawonder May 24 '19 at 22:11
  • $\begingroup$ no, the $\lambda$ is not complex $\endgroup$ – daljit97 May 24 '19 at 22:49
  • 1
    $\begingroup$ @daljit97 I adjusted my answer. See if it satisfies your conditions. $\endgroup$ – Allawonder May 24 '19 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.