Complex Analysis Homework 2

Name: Shubh Sharma Roll Number: BMC202170

Page 47

1.

$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots ,\cos z=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$

To show:

$$S_k(z)=\sin z-\sum _{i=0}^k(-1{)}^i\frac{z^{2i+1}}{(2i+1)!},C_k(z)=\cos z-\sum _{i=0}^k(-1{)}^i\frac{z^{2i}}{(2i)!}$$

have the same sign as $(-1{)}^{k+1}$ Inducting on $n$, which is the term in the considered series with the highest degree. Base case: The condititon is clear for $n=1,0$ Induction Hypothesis: The condition if true for all $n\in \{1,2,3,\cdots n-1\}$ Induction Step: If $n(2k+1)$ is odd: By the induction hypothesis, $C_k(z)$ satisfies the condition, by integrating it over $[0,z]$ we get that $S_k(z)$ also satisfies the condition. If $n(2k)$ is even: By the induction hypothesis, $S_{k-1}(z)$ satisfies the condition, by integrating it over $[0,z]$ we get $-C_k(z)$ has the same sign as $(-1{)}^k$ hence $C_k(z)$ has the same sign as $(-1{)}^{k+1}$. This completes the Induction.

2.

$$\begin{array}{r}3<\pi <2\sqrt{3}\end{array}$$ $$\cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots$$

Hence $\cos r$ is real if $r$ is real. also ${\sin }^{2}z+{\cos }^{2}z=1$ Hence for all real $z$, $-1\le \sin z,\cos z\le 1$ Derivative of $x-\sin x$ is $1-\cos x$ which is always greater than or equal to $0$, and since $\sin 0=0$, $\sin x<x\forall x>0$ Hence $\sin \frac{\pi }{6}=\frac{1}{2}⟹\frac{1}{2}<\frac{\pi }{6}⟹3<\pi$ From the previous question, we can say that

$$\sin x\le x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}$$

putting $2\sqrt{3}$ in the above polynomial returns a negative number. Hence $\sin 2\sqrt{3}<0$ so there is a root $c_0\in [0,2\sqrt{3}]$, and $c_0$ is hence and integral multiple of $\pi$ , thus $\pi \le c_0\le 2\sqrt{3}$

4.

$$\begin{array}{rl}{e}^z& =k\\ \log (k)& =\log (|k|)+i\arg (k)\\ \text{For }k& =2\\ z& =\log (2)+2ni\pi \\ \text{For }k& =-1\\ z& =i\pi +2ni\pi \\ \text{For }k& =i\\ z& =\frac{i\pi }{2}+2ni\pi \\ \text{For }k& =\frac{-i}{2}\\ z& =\log \left(\frac{1}{2}\right)+\frac{i3\pi }{2}+2ni\pi \\ \text{For }k& =-1-i\\ z& =\log \left(\sqrt{2}\right)+\frac{i5\pi }{8}+2ni\pi \\ \text{For }k& =1+2i\\ z& =\log \left(\sqrt{5}\right)+i\arctan \left(\frac{2}{\sqrt{5}}\right)+2ni\pi \end{array}$$

6.

$$\begin{array}{rlr}{2}^i& =(e^{\log 2}{)}^i=e^{i\log (2)}=\cos (\log 2)+i\sin (\log 2)& \\ i^i& =(e^{\frac{i\pi }{2}+2in\pi }{)}^i=e^{\frac{-\pi }{2}+2n\pi }& \\ (-1{)}^i& =(e^{\log -1}{)}^i=e^{\pi +2n\pi }& \end{array}$$

7.

$$\begin{array}{rl}{z}^z& =e^{z\times \log z}\\ & =e^{z\times (\log (|z|)+i\arg (z))}\\ & =e^{Re(z)\cdot \log (|z|)-Im(z)\cdot (\arg (z)+2n\pi )}\cdot e^{i(Re(z)\cdot (\arg (z)+2m\pi )+Im(z)\cdot \log (z))}\\ & =e^{Re(z)\cdot \log (|z|)-Im(z)\cdot (\arg (z)+2n\pi )}\cdot e^{i(Re(z)\cdot \arg (z)+Im(z)\cdot \log (z))}\end{array}$$

9.

Let the triangle be formed by $3$ complex numbers $a,b,c$. let $m=\frac{a+b+c}{3}$ let $f(z)=\frac{z-m}{a}$. Since $f$ is conformal, the angle between $2$ lines does not change, hence the “angles of a trianlge” do not change If $Im(b^{\prime })>0$: let angles

$$\begin{array}{r}A=\arg (c^{\prime }-a^{\prime })-\arg (b^{\prime }-a^{\prime })\\ B=\arg (a^{\prime }-b^{\prime })-\arg (c^{\prime }-b^{\prime })\\ C=\arg (b^{\prime }-c^{\prime })-\arg (a^{\prime }-c^{\prime })\end{array}$$

where $a^{\prime }=f(a),b^{\prime }=f(b),c^{\prime }=f(c)$ Then $A+B+C=\arg (b^{\prime }-c^{\prime })-\arg (c^{\prime }-b^{\prime })+\arg (c^{\prime }-a^{\prime })-\arg (a^{\prime }-c^{\prime })+\arg (a^{\prime }-b^{\prime })-\arg (b^{\prime }-a^{\prime })$ Thus $A+B+C=-\pi +\pi +\pi =\pi$

Page 72

1.

The domain for a single valued branch of $\sqrt{z}$ is the complement of $\{x:x\in \mathbb{R},x\le 0\}$, hence for $\sqrt{1+z}$ the domain would be $\{x:x\in \mathbb{R},x\le -1\}$ and for $\sqrt{1-z}$ the domain would be $\{x:x\in \mathbb{R},x\ge 1\}$ Hence, for the function $f(z)=\sqrt{1+z}+\sqrt{1-z}$ the domain would be the intersection of two which is $\{x+iy:x,y\in \mathbb{R},y\ne 0\}\cup \{x:x\in \mathbb{R},-1\le x\le 1\}$

3.

if $Ref(z)=0$ then $f(z)=iy$ and $|f(z{)}^2-1|<1$ implies $|-y^2-1|<1$ which is false for all $y\in \mathbb{R}$ hence $Ref(z)$ is never zero in the domain. Since the domain is connected $Ref(z)$ is either greater than $0$ or less than $0$ in the entire domain.

Page 78

1.

FTSOC: If there exists a transformation, let that be

$$\overline{z}=\frac{az+b}{cz+d}$$

$\overline{0}=0$ hence $b=0$ if $r\in \mathbb{R},\overline{r}=r$

$$\begin{array}{rl}r& =\frac{ar}{cr+d}\\ c{r}^2& =(a-d)r\\ cr& =a-d\end{array}$$

which is a contradiction, hence such a linear transformation does not exist

2.

If a linear transform is represented in the following way

$$Tz=\frac{az+b}{cz+d}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$

Then composition of maps will correspond to multiplication of matrices $T_1=\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)$ $T_2=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ $T_1^{-1}=\left(\begin{array}{cc}3& -2\\ -1& 1\end{array}\right)$ Then $T_1{T}_2=\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}3& 2\\ 4& 3\end{array}\right)$ $T_1{T}_2=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 2& 5\end{array}\right)$ $T_1^{-1}{T}_2=\left(\begin{array}{cc}3& -2\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}1& -2\\ 0& 1\end{array}\right)$ Hence

$$\begin{array}{ccc}{T}_1{T}_{2}z=\frac{3z+2}{4z+3}& T_2{T}_{1}z=\frac{z+2}{2z+5}& T_1^{-1}{T}_{2}z=z-2\end{array}$$

3.

$|az-aw|=|a||z-w|=|z=w|$ if $|a|=1$ Hence rotation preserves distances. Since distance is preserved $|z|=|Tz|$. Hence all points of a circle of radius $r$ around $0$ will remain on the cirlcle. Let $a=T1$ Let $b=Tz$ for some $z\in \mathbb{C}$ $|\frac{b}{a}|=|z|$ and $|z-1|=\left|\frac{b}{a}-\frac{a}{a}\right|=\left|\frac{b}{a}-1\right|$
Since the distance from $0$ and $1$ is the same for the points $z$ and $\frac{b}{a}$, $\frac{b}{a}=z$ or $\frac{b}{a}=\overline{z}$ . Hence the transformation followed by a rotation that sends 1’s image to 1 sends all points to themselves or thier relfection in the real number line. Let that transformation be $T^{\prime }z=\frac{Tz}{T1}$ For non real numbers $a=x+iy,b=z+iw$ and $T^{\prime }a=\overline{a}$ Since $T^{\prime }$ is distance preserving

$$\begin{array}{rl}|a-b|& =|Ta-Tb|\\ |(x-z)+i(y-w)|& =|(x-z)+i(-y-w^{\prime })|\end{array}$$

is true if and only if $y-w=-y-w^{\prime }$ where $w^{\prime }=w$ or $-w$ and for $w,y\ne 0$ the statement is true iff $w^{\prime }=-w$ or $T^{\prime }b=\overline{b}$ Hence if one of the points goes to its mirror image, all points go to their mirror image. so $T^{\prime }$ is either identity or reflection about the real number line. any such $T$ can be obtained by following $T^{\prime }$ with a rotation that maps $1$ to $T1$. Hence the most general distance preserving transformations are identitiy or relfection about the real numberline followed by a rotation. If the transfomation is relfection followed by rotation, it maps $r\cdot e^{i\theta }$ to $r\cdot e^{-i\theta +i\alpha }$ which can be rewritten as $r\cdot e^{-(i\theta -i\alpha )}$ which is rotation by $-\alpha$ followed by reflection. Hence any transformation that preserves distances is of the form, a rotation followed by identity or reflection along the real number line.

4.

Let the transformation be $T{z}^{\prime }=\frac{a^{\prime }z+b^{\prime }}{c^{\prime }z+d^{\prime }}$ If $a^{\prime }\ne 0$ divide the numerator and denominator by $|a^{\prime }{|}^2$

$$Tz=\frac{z+b}{cz+d}$$

putting $z=0$ we get $b/d$ is real and taking the sequence of natural numbers $Tz$ converges to $\frac{1}{c}$ Hence $c$ is also real.

Taking $b=x+iy$ and $z=-x$ we get $T(-x)=\frac{iy}{(-cx+kx)+kiy}$ Which is real iff $c=k$ so

$$Tz=\frac{z+b}{cz+cb}=1/c$$

otherwise, if $c=rd$ where $r$ is real we get

$$Tz=\frac{b}{d(rz+1)}=\frac{b/d}{rz+1}$$

which would make $\frac{b}{d},r,1$ real otherwise, We can find real numbers $a$ and $b$ such that they would make the real and imaginary parts of the denominator $0$ respectively. Hence To make the fraction always real, we $b$ would have to be both real and imaginary so $T(z)=0$.