MTH2003 Complex Analysis Practical 1

• Practical Sheet.
• Practical Solutions.

Question 1

The complex numbers $-1,-i,1+i,\frac{1}{2}+\sqrt{\frac{3}{2}i},\frac{1}{2}-\sqrt{\frac{3}{2}i}$ have corresponding exponential forms…

$$\begin{array}{rl}-1& =r{e}^{i\theta }\\ & =\boxed{e^{i\pi }}\text{by Euler’s Identity, or as a trivial/principal answer}\end{array}$$ $$\begin{array}{rl}-i& =r{e}^{i\theta }:r=\sqrt{(0{)}^2+(-1{)}^2}=1,\theta =\arctan \frac{-1}{1}=\frac{3\pi }{2}\\ & =\boxed{e^{i\frac{3\pi }{2}}}\end{array}$$ $$\begin{array}{rl}1+i& =r{e}^{i\theta }:r=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2},\theta =\arctan \frac{1}{1}=\frac{\pi }{4}\\ & =\boxed{\sqrt{2}{e}^{i\frac{\pi }{4}}}\end{array}$$ $$\begin{array}{rl}\frac{1}{2}+\frac{\sqrt{3}}{2}i& =r{e}^{i\theta }\\ & =\boxed{e^{i\frac{\pi }{3}}}\end{array}$$ $$\begin{array}{rl}\frac{1}{2}-\frac{\sqrt{3}}{2}i& =r{e}^{i\theta }\\ & =\boxed{e^{-i\frac{\pi }{3}}}\end{array}$$

Note: should keep arguments positive, e.g. $-\frac{\pi }{3}\to \frac{5\pi }{3}$.

Question 2

The complex numbers $e^{2+i\frac{\pi }{2}}$, $\frac{1}{i}$, $\frac{1}{1+i}$, $(1+i{)}^3$, and $|3+4i|$ have corresponding rectangular forms…

$$\begin{array}{rl}{e}^{2+i\frac{\pi }{2}}& =e^2{e}^{i\frac{\pi }{2}}\\ & =e^2\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\text{by Euler’s formula}\\ & =\boxed{e^{2}i}\end{array}$$ $$\begin{array}{rl}\frac{1}{i}& =\frac{1}{i}\cdot \frac{-i}{-i}\\ & =\boxed{-i}\end{array}$$ $$\begin{array}{rl}\frac{1}{1+i}& =\frac{1}{1+i}\cdot \frac{1-i}{1-i}\\ & =\frac{1-i}{2}\\ & =\boxed{\frac{1}{2}-\frac{1}{2}i}\end{array}$$ $$\begin{array}{rl}(1+i{)}^3& =1+3i+3i^2+i^3\\ & =1+3i-3-i\\ & =\boxed{-2+2i}\end{array}$$ $$|3+4i|=\boxed{5}$$

Question 3

Sketch the following regions in the complex plane, using full lines for boundaries that are included in the region and dashed lines for boundaries that are excluded

MTH2003 Complex Analysis Practical 1, page 1

Complete these diagrams.

Question 4

For any two complex numbers, $z=a+ib$ and $w=c+id$…

$$\begin{array}{rl}\overline{z+w}& =\overline{(a+ib)+(c+id)}\\ & =\overline{(a+c)+i(b+d)}\\ & =(a+c)-i(b+d)\\ & =a-ib+c-id\\ & =\bar{z}+\bar{w}\square \end{array}$$ $$\begin{array}{rl}\overline{zw}& =\overline{(a+ib)(c+id)}\\ & =\overline{ac-bd+i(ad+bc)}\\ & =ac-bd-iad-ibc\\ & =(a-ib)(c-id)\\ & =\bar{z}\bar{w}\square \end{array}$$

As always, should actually prove both ways.

Question 5

For an arbitrary complex number $z=x+iy$…

$$\begin{array}{rl}\frac{1}{2}(z+\overline{z})& =\frac{1}{2}(x+iy+x-iy)\\ & =\frac{1}{2}(2x)\\ & =x\\ & =\text{Re}(z)\square \end{array}$$ $$\begin{array}{rl}\frac{1}{2i}(z-\overline{z})& =\frac{1}{2i}(x+iy-x+iy)\\ & =\frac{1}{2i}(2iy)\\ & =y\\ & =\text{Im}(y)\square \end{array}$$

As always, should actually prove both ways.

With the results, we can show that…

$$\begin{array}{rl}2& =x+3y\\ & =\frac{1}{2}(z+\overline{z})+\frac{3}{2i}(z-\overline{z})\\ & =\frac{1}{2}(z+\bar{z}+\frac{3}{i}z-\frac{3}{i}\bar{z})\\ & =\frac{1-3i}{2}z+\frac{1+3i}{2}\bar{z}\\ & \boxed{\therefore 4=(1-3i)z+(1+3i)\bar{z}}\end{array}$$

Question 6

To find the three cube roots of $i$, we first convert to exponential form such that $i\to e^{i\frac{\pi }{2}}$. Then…

$$\begin{array}{rl}(e^{i\frac{\pi }{2}}{)}^{\frac{1}{3}}& =(e^{i(\frac{\pi }{2}+k\pi )}{)}^{\frac{1}{3}}:k=0,1,2\\ & =e^{\frac{i(\frac{\pi }{2}+k\pi )}{3}}\\ & =e^{i\frac{\pi }{6}},e^{i\frac{3\pi }{2}},e^{i\frac{5\pi }{6}}\\ & =\boxed{\frac{\sqrt{3}}{2}+\frac{1}{2}i,-\frac{\sqrt{3}}{2}+\frac{1}{2}i,-i}\end{array}$$

Question 7

Similar to question 6, we can find the following solutions (although I’ll skip to the answer using General Solution for a Complex Root Formula)…

$$\begin{array}{rl}{z}^3=4:z_k& =4^{\frac{1}{3}}\left[\cos \left(\frac{2\pi k}{3}\right)+i\sin \left(\frac{2\pi k}{3}\right)\right]:k=0,1,2\\ & =\boxed{\sqrt[3]{4},\sqrt[3]{4}\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right),\sqrt[3]{4}\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}\end{array}$$

Question 8

Given an equation $(az+b{)}^3=c$, where $a,b,c$ are real and positive constants…

$$\begin{array}{rl}az+b& =c^{1/3}\left[\cos \left(\frac{2\pi k}{3}\right)+i\sin \left(\frac{2\pi k}{3}\right)\right]:k=0,1,2\\ & =c^{\frac{1}{3}},\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)c^{\frac{1}{3}},\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)c^{\frac{1}{3}}\end{array}$$

Which can be re-arranged for…

$$\boxed{z=\frac{c^{\frac{1}{3}}-b}{a},\frac{\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)c^{\frac{1}{3}}-b}{a},\frac{\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)c^{\frac{1}{3}}-b}{a}}$$