MTH2003 Complex Analysis Tutorial 1

• Question Sheet.

Alternative conjugate notation $z^{\ast }$.

Question 1

Show that, if $z^2=(\bar{z}{)}^2,z\ne 0$, then $z$ is either purely real ($y=0$) or purely imaginary ($x=0$).

$$\begin{array}{rl}{z}^2& =(x+iy{)}^2\\ & =x^2-y^2+2ixy\end{array}$$ $$\begin{array}{rl}{\bar{z}}^2& =(x-iy{)}^2\\ & =x^2-y^2-2ixy\end{array}$$

These are only equivalent if $2ixy=-2ixy⟺xy=-xy⟺xy=0$. This is only true if $x=0$ or $y=0\square$.

Question 2

Show that for any $z,w$, $z\bar{w}+\bar{z}w=2\text{Re}(z\bar{w})$…

$$\begin{array}{rl}\text{LHS}& =(a+ib)(c-id)+(a-ib)(c+id)\\ & =(ac-iad+ibc+bd)+(ac+iad-ibc+bd)\\ & =2(ac+bd)\\ & \\ \text{RHS}& =2\text{Re}((a+ib)(c-id))\\ & =2\text{Re}(ac-iad+ibc+bd)\\ & =2(ac+bd)\\ & \\ & \text{LHS}=\text{RHS}\square \end{array}$$

Question 3

Show that if $z+\frac{1}{z}$, then either $\text{Im}(z)=0$ or $|z|=1$…

$$\begin{array}{rl}z+\frac{1}{z}& =a+ib+\frac{1}{a+ib}\\ & =a+ib+\frac{1}{a+ib}\frac{a-ib}{a-ib}\\ & =a+ib+\frac{a-ib}{a^2+b^2}\\ & =a\left(1+\frac{1}{a^2+b^2}\right)+ib\left(1-\frac{1}{a^2+b^2}\right)\\ ...& \end{array}$$

Question 4

Part a

Plotting $\text{Im}(i+\bar{z})=4$, we’ll first simplify the equation to plot it:

$$\begin{array}{rl}4& =\text{Im}(i+\bar{z})\\ & =\text{Im}(i+x-iy)\\ & =\text{Im}(x+i(1-y))\\ & =1-y\\ \therefore y=-3& \end{array}$$

Plotted, this would be a horizontal line passing through $y=-3$.

Part b

Plotting $|z-5|=6$, we’ll get a circle equation simplified such that:

$$\begin{array}{rl}|(x-5)+iy|=6⟺& (x-5{)}^2+y^2=36\end{array}$$

Or just, by inspection, see that it’ll be a circle centred at $(5,0)$ with radius $6$.

Part c

Plotting $\text{Re}(z+2)=-1$, simplified such that:

$$\begin{array}{rl}-1& =\text{Re}(x+iy+2)\\ & =x+2\\ \therefore x=-3& \end{array}$$

Which would be a vertical line at $x=-3$.

Part d

Plotting $\text{Re}(i\bar{z})=3$, simplified such that:

$$\begin{array}{rl}3& =\text{Re}(i(x-iy))\\ & =\text{Re}(ix+y)\\ & =y\\ \therefore y=3& \end{array}$$

Which would be a horizontal line at $y=3$.

Part e

Plotting $|z+i|=|z-i|$, we’ll find the points where two circles are equal, equivalent to a perpendicular bisector of the two centres of each circle.

Mathematically, this can be shown such that:

$$\begin{array}{rl}(z+i{)}^2=(z-i{)}^2& ⟺(x+iy+i{)}^2=(x+iy-i{)}^2\\ & ⟺(x+i(y+1){)}^2=(x+i(y-1){)}^2\\ & ⟺x^2+(y+1{)}^2=x^2+(y-1{)}^2\\ & ⟺x^2+y^2+2y+1=x^2+y^2-2y+1\\ & ⟺2y=-2y\\ & ⟺y=0\square \end{array}$$

Part f

Plotting $\text{Im}\left(\frac{1}{z}\right)=3$, we first simplify such that:

$$\begin{array}{rl}3& =\text{Im}\left(\frac{1}{x+iy}\right)\\ & =\text{Im}\left(\frac{x-iy}{x^2+y^2}\right)\\ & =\text{Im}\left(\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\right)\\ & =-\frac{y}{x^2+y^2}\\ & ⟺3(x^2+y^2)=-y\\ & ⟺3x^2+3y^2+y=0\\ & ⟺3x^2+3\left(y^2+\frac{1}{3}y\right)=0\\ & ⟺3x^2+3\left({\left(y+\frac{1}{6}\right)}^2-\frac{1}{36}\right)\\ & ⟺...\end{array}$$

Therefore a circle with radius $\frac{1}{6}$ with centre at $(0,-\frac{1}{6})$.