Laurent series expansion for |z|>1\lvert z\rvert >1.

I have a simple complex function like this: z+1z−1 When I expand it by its Maclaurin series: z+1z−1=z−1+2z−1=1−21−z=1−2∞∑k=0zk subject to |z|<1. I am supposed to represent this function by its Laurent series for the domain |z|>1 but for |z|>1, the function hasn’t any singularity, also there is no descending term. How to approach this problem? … Read more

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

How is this step done? |i¯z2−i2|=|z−1|2\left|\frac{i\overline{z}}{2} -\frac i2\right|=\frac{|z-1|}{2}

Absolutely everything makes sense other than what is in red. How is this step completed? Let us show that if f(z)=i¯z2 in the open disk |z|<1, thenlimz→1f(z)=i2. That the point 1 being on the boundary of the domain of f. Observe that when z is in the disk |z|\lt 1, \left|f(z)-\frac i2\right|=\color{red}{\left|\frac{i\overline{z}}{2} -\frac i2\right|=\frac{|z-1|}{2}}. Hence … 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

Holomorphic equivalent to analytic

A holomorphic function is differentiable everywhere and satisfies Cauchy-Riemann condition. Prove that a function is holomorphic if and only if it’s analytic? I have no idea how to prove this. Thanks… Answer Let $G$ be a domain and $f: G \to \mathbb{C}$ be a function who is differentiable everywhere in $G$ and satisfies Cauchy-Riemann equations, … Read more

Complex number, series

Show that 1z2=1+∞∑n=1(n+1)(z+1)n when |z+1|<1 I’m having problems to resolve this type of exercise since my book has virtually no exercises of this type, these expressions are based on Maclaurin series? Answer make the substitution w=z+1 so |z+1|<1 means |w|<1. now note that: \frac1{(1-w)^2} = \sum_{k=0}^\infty \binom{-2}{k} (-w)^k = 1 +2w+3w^2+… AttributionSource : Link , … Read more

$\langle A,B\rangle = \operatorname{tr}(B^*A)$

“define the inner product of two matrices $A$ and $B$ in $M_{n\times n}(F)$ by $$\langle A,B \rangle = \operatorname{tr}(B^*A), $$ where the {conjugate transpose} (or {adjoint}) $B^*$ of a matrix $B$ is defined by $B^*_{ij} = \overline{B_{ji}}.$ Prove that $\langle B,A \rangle = \overline{\langle A,B \rangle}$ This is left to the reader in my linear … Read more