Unlock the Secrets of Complex Numbers: 3 Essential Exercises to Master Your Math Skills

Here is a persuasive advertisement for ghostwriting services for Complex Analysis:

**Unlock the Secrets of Complex Analysis with Our Expert Ghostwriting Services**

Struggling to solve complex number problems or prove properties of linear maps? Do you need help with your complex analysis assignments or research papers?

Our team of expert ghostwriters specializes in providing top-notch writing services for complex analysis students. With our help, you can:

* Get accurate solutions to complex number problems, including finding real and imaginary parts
* Prove properties of linear maps with ease and confidence
* Receive well-researched and well-written papers that meet the highest academic standards
* Enjoy fast turnaround times and flexible deadlines to meet your needs

Our ghostwriting services offer:

* 100% original and plagiarism-free content
* Expert knowledge of complex analysis concepts and theories
* Clear and concise writing style that ensures easy understanding
* Unlimited revisions to ensure your satisfaction

Don’t let complex analysis problems hold you back. Our expert ghostwriters are here to help you succeed. Order now and get:

* A 10% discount on your first order
* A free consultation with our expert ghostwriters
* A guarantee of timely delivery and satisfaction

Don’t wait – unlock the secrets of complex analysis with our expert ghostwriting services. Order now and achieve academic success!

\[3x+2iy-ix+5y=7+5i.\]
4. Find the real and imaginary parts of the following where \(z=x+iy\):

a) \(\frac{1}{z^{2}}\) b) \(\frac{1}{3z+2}\).
5. If \(a\) is real and \(z\) is complex, prove that \(\ \operatorname{Re}(az)=a\operatorname{Re}(z)\\) and that \(\operatorname{Im}(az)=a\operatorname{Im}(z)\). Generally, show that \(\operatorname{Re}:\mathbb{C}\longrightarrow\mathbb{R}\) is a linear map, that is,

\[\operatorname{Re}(az+bw)=a\operatorname{Re}z+b\operatorname{Re}w\]

for \(a,b\) real, \(z,w\) complex.
6. Prove that, for any integer \(k\),

\[i^{4k}=1,\qquad i^{4k+1}=i,\qquad i^{4k+2}=-1,\qquad i^{4k+3}=-i.\]

Show how this result gives a formula for \(i^{n}\) for all \(n\) by writing \(n=4k+j\), \(0\leq j\leq 3\).
7. Sketch the sets of points that satisfy the following equations:\[\begin{array}{ccccc}\text{a) }|z|=2,&\text{b) }|z-i|=|z+i|,&\text{c) }|2z-4i|<1,\\ &\text{d) }2>\operatorname{Re}z>-3,&\text{e) }(\operatorname{Im}z)^{2}\leq \operatorname{Re}z.\end{array}\]

8. Show that the field of complex numbers \(\mathbb{C}\) is not an ordered field.

9. Let \(p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\) be a polynomial with real coefficients \(a_{0},a_{1},a_{2},\ldots,a_{n}\). Prove that if \(z\) is a solution of the equation \(p(x)=0\), then so is \(\overline{z}\).

10. Solve \(z^{8}=1\) for \(z\).

11. Find each of the indicated roots and locate them graphically:

\[\text{a) }(-1+i)^{\frac{1}{3}},\qquad\qquad\text{b) }(-2\sqrt{3}-2i)^{\frac{1}{4}}.\]

12. Express \(\cos 3\theta\) in terms of \(\cos\theta\) and \(\sin\theta\) using de Moivre’s formula.

13. For \(z\neq 1\) verify the identity

\[1+z+z^{2}+\cdots+z^{n}=\frac{1-z^{n+1}}{1-z}.\]

14. Use the problem 13 above to establish that

\[1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta=\frac{1}{2}+\frac{\sin (n+\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta}\]

for \(0<\theta<2\pi\). This result is known as _Lagrange's identity_ and is useful in the theory of Fourier series.

### 3.2 Holomorphic Functions

Complex analysis is the study of holomorphic (sometimes called analytic) functions. Holomorphic means almost the same thing as differentiable, but there is an important distinction between the two concepts caused by the role played by open sets. In studying differentiable functions of a real variable, we consider functions defined on intervals. For a complex-valued function, we need to select the domains of our definition as an open set.

**Definition 103**.: Let \(f:G\subset\mathbb{C}\to\mathbb{C}\) be a function from an open set \(G\) of complex numbers to the complex numbers. We say \(f\) is _differentiable_ at \(z\in G\) if

\[\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\]

exists (independently of the manner in which \(h\) approaches \(0\)) for each \(z\in G\). This limit is denoted by \(f^{\prime}(z)\) and is called the complex derivative.

**Definition 104**.: A complex-valued function \(f\) which is differentiable at every point of an open set \(G\) is said to be _holomorphic_ in \(G\) and we use \(H(G)\) to denote the set of all functions holomorphic in \(G\).

A complex-valued function \(f\) is said to be _holomorphic at a point_\(a\in\mathbb{C}\) if there exists \(r>0\) such that \(f\) is defined and is holomorphic in \(D(a;r)\), where \(D(a;r)\) is the open disc defined as \(D(a;r):=\{z\in\mathbb{C}:|z-a|

Note that \(G\) is open, thus, given \(z\in G\) there exists \(r>0\) such that \(D(z,r)\subset G\). That is, \(z+h\in G\) whenever \(|h|

\[\lim_{h\to 0}\dfrac{\overline{(z+h)}-\overline{z}}{h}=\lim_{h\to 0}\dfrac{ \overline{h}}{h}=\begin{cases}1&\text{if $h$ is real}\\ -1&\text{if $h$ is purely imaginary}.\end{cases}\]

The idea of restricting \(h\) to real and imaginary values can be used to obtain a necessary but not sufficient condition for differentiability.

### Cauchy-Riemann Equations

In this section we will see that a differentiable function gives rise to a set of wonderful partial differential equations called the Cauchy-Riemann equations.

**Theorem 121**.: _Let \(f\) be a complex-valued function defined on an open set \(G\) and let \(f\) be differentiable at \(z=x+iy\in G\). Let \(f(z)=u(x,y)+iv(x,y)\). Then \(u\) and \(v\) have first-order partial derivatives (denoted by \(u_{x}\), \(u_{y}\), \(v_{x}\), \(v_{y}\)) and these partial derivatives satisfy the Cauchy-Riemann equations:_

\[u_{x}=v_{y}\text{, }v_{y}=-v_{x}.\]

Proof

: Since \(f\) is differentiable at \(z\in G\), \(f^{\prime}(z)=\lim_{h\to 0}\dfrac{f(z+h)-f(z)}{h}\) exists. We calculate \(f^{\prime}(z)\) in two different ways, restricting \(h\) to be real and purely imaginary.

Case 1: Suppose \(h\) is real. Then writing

\[f(z+h) =u(z+h)+iv(z+h)\] \[=u(x+h,y)+iv(x+h,y)\]in the definition of \(f^{\prime}(z)\), we have

\[f^{\prime}(z) =\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\] \[=\lim_{h\to 0}\left[\frac{u(x+h,y)-u(x,y)}{h}+i\frac{v(x+h,y)-v(x,y) }{h}\right]\] \[=u_{x}+iv_{x}.\]

Case 2: Suppose \(h\) is purely imaginary (i.e., \(h=ik\), \(k\in\mathbb{R}\)). Then,

\[f(z+h)=f(z+ik)=u(x,y+k)+iv(x,y+k)\]

and

\[f^{\prime}(z) =\lim_{\begin{subarray}{c}h\to 0\\ h=ik\end{subarray}}\left[\frac{u(x,y+k)-u(x,y)}{ik}+i\frac{v(x,y+k)-v(x,y)}{ik}\right]\] \[=\frac{1}{i}u_{y}+v_{y}.\]

Equating the two expressions for \(f^{\prime}(z)\)

\[u_{x}+iv_{x}=-iu_{y}+v_{y}\]

and equating real and imaginary parts we obtain

\[u_{x}=v_{y}\] \[u_{y}=-v_{x}.\qed\]

The converse of Theorem 121 is false as seen in the following example:

**Example 98**.: Let \(f(z)=f(x+iy)=0\) if both \(x\) and \(y\) are zero and \(f(z)=1\) if neither \(x\) nor \(y\) is zero. Then \(u_{x}=u_{y}=v_{x}=v_{y}=0\). Thus, the Cauchy-Riemann equations hold. However, \(f(z)\) is not differentiable at \(0\), because

\[\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}\]

fails to exist if \(h\to 0\) along the ray \(\arg z=\pi/4\).

Note that the contrapositive of Theorem 121 is useful for proving non-differentiability, as shown in the following example.

**Example 99**.: To show the complex function \(f(z)=x+4iy\) is not differentiable at any point \(z\), we identify the real and imaginary parts of \(f(z)\):

\[\begin{array}{ccc}u(x,y)=x,&v(x,y)=4y,\\ u_{x}=1,&&v_{x}=0,\\ u_{y}=0,&&v_{y}=4.\end{array}\]