Find the real and imaginary part of $$f(z)=(−3y^2+6ixy+3x^2)e^{2(iy+x)}$$
I came up with: \begin{align}Re &= e^{2x}(-3y^2\cos(y)-6xy\sin(y)+3x^2\cos(y))\\Im&= ie^{2x}(-3y^2\sin(y)+6xy\cos(y)+3x\sin(y))\end{align}
Answer
Yes we can use that
$$e^{2(iy+x)}=e^{2x}(\cos(2y)+i\sin(2y))$$
and therefore
$$\Re (f(z))=(-3y^2+3x^2)e^{2x}\cos(2y)-6xye^{2x}\sin(2y)$$
$$\Im (f(z))=6xye^{2x}\cos(2y)+(-3y^2+3x^2)e^{2x}\sin(2y)$$
Recall that real part and imaginary parts are real numbers.
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