Prove 1z2=∑n≥0(−1)n(n+1)(z−1)n\frac{1}{z^2}=\sum\limits_{n\ge0}(-1)^n(n+1)(z-1)^n

Prove that for any complex number z such that |z−1|<1, one has: 1z2=∑n≥0(−1)n(n+1)(z−1)n What I’ve done; 1z2=(1z)2=(11−(1−z))2=(∑k≥0(1−z)k)2=∑n≥0∑i+j=k(1−z)i(1−z)j =∑n≥0∑i+j=k(−1)k(z−1)i(z−1)j but is the number of cases i+j=k not always even ? and are my steps correct ? Of course you can also give another hint. Answer First replace i+j=k by i+j=n, the first steps are correct but … Read more

Finding a Laurent Series involving two poles

Find the Laurent Series on the annulus 1<|z|<4 for R(z)=z+2(z2−5z+4) So I am having a few issues with this. I know there are two poles in this problem particulaly z=1 and z=4, so if I factor out I get it into a form as: z+2(z−1)(z−4) and here is where it get’s a little hazy. I … Read more

Simplify Im(az+bcz+d)Im \left(\frac{az+b}{cz+d}\right)

Let z∈H, where H denotes the half plane H={z∈C:Im(z)>0}. Let f(z)=az+bcz+d which is called a Mobius Transformation, and let ad−bc>0. I want to show that Im(f(z))>0. Following this solution, I should use Im(z)=z−¯z2i. Applying this formula, I get Im(f(z))=Im(az+bcz+d)=az+bcz+d−¯(az+bcz+d)2i but I am not sure how to show that this equals ad−bcc2+d2. Is there a nice … Read more

Why $F(z) = |z|^2$ is holomorphic nowhere?

I am self-studying basic complex analysis, and am slightly confused as to how to show that $F(z) = |z|^2$ is holomorphic nowhere. A necessary and sufficient condition for the holomorphism of $F(z)$ is that $F(z)$ is independent of $\overline{z}$. That is, we require: $$\frac{\partial F}{\partial \overline{z}}=0$$ We note that we have $F(z) = |z|^{2} = … Read more

Find the poles of f(z)=11+zwf(z)=\frac 1{1+z^w} for w>1w \gt 1

I am trying to use contour integration on the following integrand between 0 and ∞, however I am not sure how to go about finding the poles for it: f(z)=11+zw,w∈Z:w>1 Consider the denominator equal to zero: 1+zw=0 ⇒zw=−1 ⇒z=−11w≡w√z How would I go about determining the types of poles we would have for f(z) given … Read more

System of equations with complex numbers-circles

The system of equations \begin{align*} |z – 2 – 2i| &= \sqrt{23}, \\ |z – 8 – 5i| &= \sqrt{38} \end{align*} has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$. So far I have gotten the two original equations to equations of circles, $(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$. From here how do … Read more

Proving |z1z2|=|z1||z2||z_1z_2|=|z_1||z_2| using exponential form

Problem: Prove |z1z2|=|z1||z2| where z1,z2 are Complex Numbers. I tried to solve this using the exponential form of a Complex Number. Assuming z1=r1eiθ1 and z2=r2eiθ2, I got |z1z2|=|r1eiθ1×r2eiθ2|=|r1r2ei(θ1+θ2)| I cannot proceed further. Any help would be appreciated. Answer Hint: ∀θ∈R we have. |eiθ|=|cosθ+isinθ|=cos2θ+sin2θ=1 AttributionSource : Link , Question Author : User1234 , Answer Author : … Read more

Solve complex equation 5|z|^3+2+3 (\bar z) ^6=05|z|^3+2+3 (\bar z) ^6=0

I’m stuck in trying to solve this complex equation 5|z|^3+2+3 (\bar z)^6=0 where \bar z is the complex conjugate. Here’s my reasoning: using z= \rho e^{i \theta} I would write 5\rho^3+ 2 + 3 \rho^6 e^{-i6\theta} = 0 \\ 5\rho^3+ 2 + 3 \rho^6 (\cos(6 \theta) – i \cdot \sin(6 \theta)) = 0 \\ from … Read more