2

When solving this equation, I reach a point where

$$e^{2z} =\frac{1-8i}{65}$$

Now I'm pretty sure this is correct since this is what the memo has as well.

But what I would like to know is why I cannot simply let $z= a+bi$ and then solve as follows:
$$e^{2a+2bi} = \frac{1-8i}{65}$$

so that $ e^{2a} = 1/65 \rightarrow a = 0.5\ln(1/65)$ and then solve for $b$ in the same manner but the memo gets something different. Where am I going wrong?

Null
  • 1,332
dan
  • 21

1 Answers1

1

Dude you can set z=a+ib but yourproblem is that u made a mistake afterwards. The real part of the LHS is NOT e^2a. The LHS is "in polar form" so to speak. I suggest you equate the modulus of the LHS i.e. e^2a with the modulus of the RHS and same for the angles.

Nikos Δr
  • 141