Real and Imaginary part of f(z) where z=x-iy

Question

$$f(z)= exp(\frac{1}{z})$$ hello, i am trying to check if f(z) is analytic using cauchy-reimann equations. please help i am stuck trying to derive the real and imaginary parts. thank you

Answer 1

Notice: $$z=x-jy$$ $$\frac1z=\frac{1}{x-jy}\frac{x+jy}{x+jy}=\frac{x+jy}{x^2+y^2}$$ Now we split it up, since: $$e^{a+b}=e^ae^b$$ $$f(z)=\exp\left(\frac{x}{x^2+y^2}\right)\exp\left(\frac{jy}{x^2+y^2}\right)$$ we also know that: $$e^{ja}=\cos(a)+j\sin(a)$$ and so: $$f(z)=\exp\left(\frac x{x^2+y^2}\right)\left[\cos\left(\frac y{x^2+y^2}\right)+j\sin\left(\frac y{x^2+y^2}\right)\right]$$ finally we can break this up to get: $$\Re(f(z))=\exp\left(\frac x{x^2+y^2}\right)\cos\left(\frac y{x^2+y^2}\right)$$ $$\Im(f(z))=\exp\left(\frac x{x^2+y^2}\right)\sin\left(\frac y{x^2+y^2}\right)$$

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The cauchy-Riemann equations state that if: $$f(z)=f(x+jy)=u(x,y)+jv(x,y)$$ the function must satisfy the equations: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ for our equation we have: $$u(x,y)=\exp\left(\frac x{x^2+y^2}\right)\cos\left(\frac y{x^2+y^2}\right)$$ $$v(x,y)=\exp\left(\frac x{x^2+y^2}\right)\sin\left(\frac y{x^2+y^2}\right)$$ so now simply calculate these derivatives.

Answer 2

HINT

If $z = x - iy$ then $$ \frac1z = \frac{1}{x-iy} = \frac{x+iy}{x^2+y^2}, $$ and $x^2+y^2 \in \mathbb{R}$.

Can you complete this now?