PYQs

From Cauchy Integral Formula, we have

\[f^{n}(a)=\frac{n !}{2 \pi i} \int_{C} \frac{f(z) d z}{(z-a)^{n+1}}\]

Put \(a=-1\), \(n=3\)

\[f^{3}(-1)=\frac{3 !}{2 \pi i} \int_{C} \frac{f(z) d z}{(z+1)^{4}} \cdots (i)\]

Take \(f(z)=e^{3 z}\), then \(f^{n}(z)=3^{n} e^{3 z}\)

\[\therefore (3)(-1)=3^{3} e^{3}=\frac{27}{e^{3}}\]

\(\therefore\) from \((i)\)

\[\frac{27}{e^{3}}=\frac{3 !}{2 \pi i} \int \frac{e^{3 z} d z}{(z+1)^{4}}\] \[\Rightarrow \int_{c} \frac{e^{3 z} d z}{(z+1)^{4}}=\frac{9 \pi i}{e^{3}}\]

Try yourself

We can assume without loss of generality that the zeros of \(P(z)\) lie in the half plane Re \(z<0 .\) Let \(P(z)=\prod_{j=1}^{n}\left(z-\alpha_{j}\right)\) where \(\alpha_{j}=x_{j}+i y_{j}, x_{j}<0\)

If \(\operatorname{Re} z \geq 0,\) then \(P(z) \neq 0\) and

\[\begin{aligned} \frac{P^{\prime}(z)}{P(z)} &=\sum_{j=1}^{n} \frac{1}{z-\alpha_{j}} \\ &=\sum_{j=1}^{n} \frac{1}{x-x_{j}+i\left(y-y_{j}\right)} \\ &=\sum_{j=1}^{n} \frac{x-x_{j}-i\left(y-y_{j}\right)}{\left(x-x_{j}\right)^{2}+\left(y-y_{j}\right)^{2}} \end{aligned}\]

Since \(x_{j}<0,1 \leq j \leq n,\) it follows that

\[\operatorname{Re}\left(\frac{P^{\prime}(z)}{P(z)}\right)=\sum_{j=1}^{n} \frac{x-x_{j}}{\left(x-x_{j}\right)^{2}+\left(y-y_{j}\right)^{2}}>0\]

whenever Re \(z=x \geq 0 .\) Thus \(\frac{x^{\prime} \gamma}{P(z)}\) and therefore \(P^{\prime}(z)\) has no zeros in the right half plane Re \(z \geq 0\). Hence all zeros of \(P^{\prime}(z)\) lie in the same half plane in which the zeros of \(P(z)\) lie.

Let \(G(z)=f(z)-g(z),\) then \(G\left(\frac{1}{n}\right)=0\) for \(n=1,2, \ldots .\) We shall show that \(G(z) \equiv 0\) for \(z \in \mathbb{C}\) which would prove the result.

Let \(G(z)=\sum_{n=0}^{\infty} a_{n} z^{n}\) be the power series of \(G(z)\) with center 0 and radius of convergence \(R,\) clearly \(R>0 .\) We shall now prove that \(a_{n}=0\) for every \(n\) If \(a_{n} \neq 0\) for some \(n,\) let \(a_{k}\) be the first non-zero coefficient. Then \(G(z)=z^{k}\left(a_{k}+a_{k+1} z+\ldots\right)=z^{k} H(z)\) Clearly \(H(z)\) is analytic in \(\vert z\vert <R,\) and \(H(0) \neq 0 .\) We now claim that \(H(z) \neq 0\) in a neighborhood \(\vert z\vert < \delta\) of \(0 .\) Let \(\epsilon=\frac{\vert H(0)\vert }{2},\) then continuity of \(H(z)\) at \(z=0\) implies that there exists a \(\delta>0\) such that \(\vert z\vert <\delta \Rightarrow\vert H(z)-H(0)\vert <\epsilon\) or \(\vert H(0)\vert -\epsilon<\vert H(z)\vert <\vert H(0)\vert +\epsilon\) for \(\vert z\vert <\delta .\)

Thus \(\vert H(z)\vert >\frac{\vert H(0)\vert }{2}>0\) for \(\vert z\vert <\delta .\) Consequently, \(G(z) \neq 0\) for any \(z\) in \(0<\vert z\vert <\delta .\) But this is not possible, as \(\vert z\vert <\delta\) contains all but finitely many \(\frac{1}{n},\) at which \(G(z)\) vanishes. Thus our assumption that \(a_{n} \neq 0\) for some \(n\) is false, thus \(G(z) \equiv 0\) in \(\vert z\vert <R\)

Let \(z^{\prime}\) be any point in \(\mathbb{C},\) and let \(r(t), a \leq t \leq b\) be a continuous curve joining 0 and \(z^{\prime}\). Using uniform continuity of \(r(t),\) we get a partition \(a=t_{0}<t_{1}<\ldots<t_{n}=b\) of \([a, b]\) such that \(r\left(t_{0}\right)=0, r\left(t_{1}\right)=z_{1}, \ldots, r\left(t_{n}\right)=r(b)=z^{\prime},\) and \(\left\vert z_{j}-z_{j-1}\right\vert < R\).

Now the disc \(K_{0}=\vert z-0\vert <R\) contains \(z_{1}\), the center of disc \(K_{1}=\left\vert z-z_{1}\right\vert < R\). since \(G\left(z_{1}\right)=0\) as \(z_{1} \in K_{0} \cap K_{1},\) and \(K_{0} \cap K_{1}\) contains a sequence of points \(y_{n}\) such that \(y_{n} \rightarrow z_{1}\) and \(G\left(y_{n}\right)=0,\) we can prove as before that \(G(z) \equiv 0\) in \(K_{1}\). Proceeding in this way, in \(n\) steps we get \(G(z) \equiv 0\) in \(K_{n},\) or \(G\left(z^{\prime}\right)=0 .\) since \(z^{\prime}\) is an arbitrary point of \(\mathbb{C},\) we get \(G(z) \equiv 0\) in \(\mathbb{C}\).

\(\left\vert \frac{1}{n^{z}}\right\vert =\left\vert \frac{1}{n^{x} \cdot n^{i y}}\right\vert =\left\vert \frac{1}{n^{x}}\right\vert \quad \because\left\vert \frac{1}{n^{i y}}\right\vert =\left\vert \frac{1}{e^{i y \log n}}\right\vert =1\) since \(\sum_{n=1}^{\infty} \frac{1}{n^{x}}\) converges for \(x>1,\) it follows that \(\sum_{n=1}^{\infty} n^{-z}\) converges absolutely for \(\operatorname{Re} z>1\). If Re \(z \geq 1+\epsilon,\) then \(\frac{1}{n^{x}} \leq \frac{1}{n^{1+\epsilon}}\) and \(\sum_{n=1}^{\infty}\left\vert n^{-z}\right\vert \leq \sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon}}\) for Re \(z \geq 1+\epsilon\). Weierstrass’ M-test gives that the given series converges uniformly and absolutely for \(\operatorname{Re} z \geq 1+\epsilon\)

Hint: Take \(f(z)= u+ i u^2\) and apply CR equations.

In the final step, we get,

\(u_x=0\), \(u_y=0\), \(v_x=0\), \(v_y=0\)

Hence, \(u=c_1\) and \(v=c_2\)

\(\implies\) \(f(z) = c_1+ic_2\)is constant.

Let, \(f(z)=u+i v\) \(u=(x-1)^{3}-3 x y^{2}+3 y^{2}\) \(u_{x}=3(x-1)^{2}-3 y^{2}\) \(u_{xx}=6(x-1) \cdots (1)\) \(u y=-6 x y+6 y\) \(u_{yy}=-6 x+6=-6(x-1) \cdots (2)\)

\(\therefore \quad u_{xx}+u_{y y}=0\) Hence, function \(u\) is harmonic.

We can use Milne’s method to find harmonic conjugate of \(u\).

\(\phi_{1}(z,0)=\left( \frac{\partial u}{\partial x}\ \right)_{x=z,y=0}=3(z-1)^{2}\) \(\phi_{2}(z,0)=\left( \frac{\partial u}{\partial y}\ \right)_{x=z,y=0}=0\) \(f(z)=\int\left[\phi_{1}(z, 0)-i \phi_{2}(z, 0)\right] d z+c\) \(=\int 3(z-1)^{2} d z+c\) \(=(z-1)^{3}+c\)

\[\therefore f(z)=(z-1)^{3}+c\] \[\begin{aligned} f(z) &=[(x-1)+i y]^{3}+c \\ &=(x-1)^{3}-i y^{3}+3(x-1) i y[(x-1)+i y] \\ &=(x-1)^{3}-3(x-1) y^{2}-i y^{3}+3 i(x-1)^{2} y \\ &=(x-1)^{3}-3(x-1) y^{2}+i\left[3(x-1)^{2} y-y^{3}\right] \\ & \therefore v(x, y)=3(x-1)^{2} y-y^{3} \end{aligned}\]

If \(f(z)\) is analytic in \(\mathcal{R},\) then the Cauchy-Riemann equations

\[\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \cdots (i)\]

and

\[\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} \cdots (ii)\]

are satisfied in \(\mathcal{R}\) and g \(u\) and \(v\) have continuous second partial derivatives. We can differentiate both sides of \((i)\) with respect to \(x\) and \((ii)\) with respect to \(y\) to obtain

\[\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} v}{\partial x \partial y}\]

and

\[\frac{\partial^{2} v}{\partial y \partial x}=-\frac{\partial^{2} u}{\partial y^{2}}\]

from which

\[\frac{\partial^{2} u}{\partial x^{2}}=-\frac{\partial^{2} u}{\partial y^{2}} \quad \text { or } \quad \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0\]

Similarly, by differentiating both sides of \((i)\) with respect to \(y\) and \((ii)\) with respect to \(x\), we get \(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}=0\)

Let \(f\) be the function \(f=u+i v\) Given: \(v=x^{3}-3 x y^{2}+2 y\) Therefore, \(\begin{array}{l} v_{x}=3 x^{2}-3 y^{2}, v_{x x}=6 x \cdots (i) \\ v_{y}=-6 x y+2, v_{y y}=-6 x \cdots (ii) \end{array}\) Now, \(v_{x x}+v_{y y}=+6 x-6 x=0\) \(\Rightarrow v\) is a harmonic function. Let \(u\) be harmonic conjugate of \(v\). By Cauchy-Riemann equations, \(\begin{aligned} u_{x} &=v_{y} \Rightarrow u_{x}=-6 x y+2 \\ \frac{\partial u}{\partial x} &=-6 x y+2 \end{aligned}\) \(\Rightarrow\) \(\int d u=\int(-6 x y+2) d x\) \(\Rightarrow u=-3 x^{2} y+2 x+g(y)\), where \(g\) is an arbitrary function of \(y\) \(\cdots (iii)\) Also, \(u_{y}=-v_{x} \Rightarrow u_{y}=-\left(3 x^{2}-3 y^{2}\right)\) \(\Rightarrow\) \(\frac{\partial u}{\partial y}=-3 x^{2}+3 y^{2}\) \(\Rightarrow\) \(\int d u=\int\left(-3 x^{2}+3 y^{2}\right) d y\) \(\Rightarrow u=-3 x^{2} y+y^{3}+n(x),\) where \(n(x)\) is an arbitrary fucntion of \(x \ldots\) (iv) From \((iii)\) and \((iv)\) \(u=-3 x^{2} y+2 x+y^{3}+c, \text { where } c \text { is a constant }\) So, \(f=\left(-3 x^{2} y+2 x+y^{3}+c\right)+i\left(x^{3}-3 x y^{2}+2 y\right)\)

Let \(g(z)=z^{7}-5 z^{3}, f(z)=12,\) then i) On \(\vert z \vert =1, \vert g(z) \vert \leq\left \vert z^{7}\right \vert +5\left \vert z^{3}\right \vert =6<12= \vert f(z) \vert\). ii) Both \(g(z)\) and \(f(z)\) are analytic on and within \(\vert z \vert <1\) iii) Both \(g(z)\) and \(f(z)\) have no zeros on \(\vert z \vert =1\)

By Rouche’s theorem, \(f(z)+g(z)=z^{7}-5 z^{3}+12\) and \(f(z)\) have the same number of zeros inside \(\vert z \vert =1 .\) But \(f(z)=12\) has no zeros anywhere and in particular in the region \(\vert z \vert <1\) therefore \(z^{7}-5 z^{3}+12\) has no zeros inside the unit circle. Now we take \(g(z)=12-5 z^{3}, f(z)=z^{7}\)

i) On \(\vert z \vert =2, \vert g(z) \vert \leq 12+5\left \vert z^{3}\right \vert =52<2^{7}= \vert f(z) \vert\). ii) Both \(g(z)\) and \(f(z)\) are analytic on and within \(\vert z \vert <2\)

Therefore by Rouche’s theorem, \(g(z)+f(z)=z^{7}-5 z^{3}+12\) and \(f(z)\) have the same number of zeros inside \(\vert z \vert =2 .\) since \(f(z)=z^{7}\) has 7 zeros \((z=0\) is a zero of order 7 of \(f(z))\) inside \(\vert z \vert =2,\) the given polynomial has seven zeros inside \(\vert z \vert =2\) i.e. all its zeros lie inside \(\vert z \vert =2\) since \(z^{7}-5 z^{3}+12\) has no zeros inside and on \(\vert z \vert =1,\) therefore all zeros lie in the ring \(1< \vert z \vert <2\).

Let \(F(z)=(1+i) f(z)=(1+i)(u+i v)=(u-v)+i(u+v) .\) Now \(\frac{\partial^{2}(u-v)}{\partial x^{2}}+\frac{\partial^{2}(u-v)}{\partial y^{2}}=e^{x}(\cos y-\sin y)+e^{x}(-\cos y+\sin y)=0\) Let \(F(z)=U+i V,\) where \(U=u-v\) is harmonic. If \(f(z)\) is analytic, then so is \(F(z)\) and \(\frac{d F}{d z}=\frac{\partial U}{\partial x}+i \frac{\partial V}{\partial x}=\frac{\partial U}{\partial x}-i \frac{\partial U}{\partial y}\) as \(\frac{\partial V}{\partial x}=-\frac{\partial U}{\partial y}\) by the Cauchy-Riemann equations. Thus \(\begin{aligned} F^{\prime}(z) &=e^{x}(\cos y-\sin y)-i e^{x}(-\sin y-\cos y) \\ &=e^{x}(\cos y+i \sin y)+i e^{x}(\cos y+i \sin y) \\ &=(1+i) e^{x} \cdot e^{i y}=(1+i) e^{z} \end{aligned}\) Thus \(F(z)=(1+i) e^{z}\) and \(f(z)=e^{z},\) which is the required function.