Functions

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Question 1

Evaluate the following functions at the indicated points:

a) $f(x) = z^{3}+2z$ at $z = i$

(1)
\begin{equation} f(i) = i^{3}+2i \end{equation}
(2)
\begin{equation} = -i + 2i \end{equation}
(3)
\begin{equation} = i \end{equation}

b) $f(z) = e^{z}$ at $z=2+\frac{\pi}{3}i$

(4)
\begin{align} f(i) = e^{2+\frac{\pi}{3}i} \end{align}
(5)
\begin{align} = e^{2}e^{\frac{\pi}{3}i} \end{align}
(6)
\begin{align} = e^{2}(\cos (\frac{\pi}{3})+i \sin (\frac{\pi}{3})) \end{align}
(7)
\begin{align} = e^{2}\cos (\frac{\pi}{3})+ie^{2} \sin (\frac{\pi}{3}) \end{align}

c) $f(z)= z+\frac{1}{z}$ at $z= 1 +2i$

(8)
\begin{align} 1 +2i + \frac{1}{1+2i} *\frac{(1-2i)}{(1-2i)} \end{align}
(9)
\begin{align} 1 +2i + \frac{1-2i}{1+4} \end{align}
(10)
\begin{align} 1 +2i + \frac{1-2i}{5} \end{align}
(11)
\begin{align} \frac{5+10i}{5}+\frac{1-2i}{5} \end{align}
(12)
\begin{align} \frac{6+8i}{5} \end{align}

Question 2

Find the real and imaginary parts in terms of rectangular coordinates x and y.
In each case, set $w=u+i v$ and $z=x+i y$.

a) $w=6z+5$

(13)
\begin{eqnarray} u+i v & = & 6z+5\\ & = & 6(x+iy)+5\\ & = & 6x+6iy+5 \end{eqnarray}
(14)
\begin{Bmatrix} u & = & 6x+5 \\ v & = & 6y \end{Bmatrix}

b) $w = z^3-2z$

(15)
\begin{eqnarray} u + i v & = & (x+iy)^3-2(x+iy)\\ & = & (x+iy)(x+iy)(x+iy)-2(x+iy)\\ & = & (x^2+2xiy+i^2y^2)(x+iy)-2(x+iy)\\ & = & (x^3+2x^2iy+i^2y^2x+x^2iy+2xi^2y^2+i^3y^3)-2(x+iy)\\ & = & (x^3+3x^2iy+3xi^2y^2+i^3y^3)-2(x+iy)\\ & = & (x^3+3x^2iy-3xy^2-iy^3)-2(x+iy)\\ & = & (x^3+3x^2iy-3xy^2-iy^3-2x-2iy \end{eqnarray}
(16)
\begin{Bmatrix} u & = & x^3-3xy^2-2x \\ v & = & 3x^2y-y^3-2y \end{Bmatrix}

c) $w= \frac{1}{z}$

(17)
\begin{eqnarray} u + i v & = & \frac{1}{x+iy} \\ & = & \frac{(x-iy)}{(x+iy)(x-iy)} \\ & = & \frac{x-iy}{x^2-xiy+xiy-i^2y^2}\\ & = & \frac{x-iy}{x^2+y^2} \\ & = & \frac{x}{x^2+y^2}- \frac{iy}{x^2+y^2} \end{eqnarray}
(18)
\begin{Bmatrix} u & = & \frac{x}{x^2+y^2} \\ v & = & -\frac{y}{x^2+y^2} \end{Bmatrix}

$w=e^z$

(19)
\begin{eqnarray} u+i v & = & e^{x+iy} \\ & = & e^x e^{iy}\\ & = & e^x(\cos y+i \sin y) \end{eqnarray}
(20)
\begin{Bmatrix} u & = & e^x \cos y \\ v & = & e^x \sin y \end{Bmatrix}

Question 3

find the real and imaginary parts of the following functions in terms of polar coordinates r and $\theta$

3. a)
$w=z^3$

(21)
\begin{align} z^3=\left ( re^{i\theta} \right )^3 \end{align}
(22)
\begin{align} =r^3e^{i3\theta}=r^3\left ( cos3\theta+isin3\theta \right ) \end{align}

real and imaginary parts are:

(23)
\begin{align} u=r^3cos3\theta \end{align}
(24)
\begin{align} v=r^3sin3\theta \end{align}

3. b)
$w=z+\frac{1}{z}$

(25)
\begin{align} z+\frac{1}{z}=re^{i\theta}+\frac{1}{re^{i\theta}} \end{align}
(26)
\begin{align} =rcos\theta +irsin\theta+\frac{1}{rcos\theta+irsin\theta} \end{align}

rationalizing the denominator of the fraction:

(27)
\begin{align} \left ( \frac{1}{rcos\theta+irsin\theta} \right )\left ( \frac{rcos\theta-irsin\theta}{rcos\theta-irsin\theta} \right )=\frac{rcos\theta-irsin\theta}{r^2cos^2\theta-ir^2cos{\theta}sin{\theta}+ir^2cos{\theta}sin{\theta}-i^2r^2sin^2\theta} \end{align}
(28)
\begin{align} =\frac{rcos\theta-irsin\theta}{r^2cos^2\theta+r^2sin^2\theta}=\frac{r\left ( cos\theta-isin\theta \right )}{r^2\left ( cos^2\theta+sin^2\theta \right )}=\frac{cos\theta-isin\theta}{r} \end{align}

back to original problem:

(29)
\begin{align} =r\left ( cos\theta+isin\theta \right )+\frac{cos\theta-isin\theta}{r}=\frac{r^2cos\theta+ir^2sin\theta+cos\theta-isin\theta}{r} \end{align}

real and imaginary parts are:

(30)
\begin{align} u=\frac{r^2cos\theta+cos\theta}{r} \end{align}
(31)
\begin{align} v=\frac{r^2sin\theta-sin\theta}{r} \end{align}

Question 5

Find the image of the sector $0\leq r\leq 2, 0\leq \theta \leq \frac{\pi }{4}$ under the following the mapping w=z+(3+2i).

assignment2%285A%29.png
assignement2%285B%29.png

Question 6

Find the image of the unit square $0\leq x\leq 1, 0\leq y\leq 1,$ under the mapping $w=(1+i)z$.

A) Use the (x,y) coordinates of points A,B,C, & D on the z-plane, use their coordinates (x,y) to find the coordinates of the mapping of the w-plane:

(32)
\begin{eqnarray} u+i v & = & (1+i)(x+iy) \\ & = & x+iy+ix-i^2y \\ & = & (x-y) + i (x+y) \end{eqnarray}
(33)
\begin{Bmatrix} u & = & x-y \\ v & = & x+y \end{Bmatrix}

Now to find the w-plane coordinates, use the u and v equations found above.

Coordinate Name Z-Plane Algebra W-Plane
A (0,0) (0-0,0+0) (0,0)
B (0,1) (0-1,0+1) (-1,1)
C (1,0) (1-0,1+0) (1,1)
D (1,1) (1-1,1+1) (0,2)
MappingSquare.png

B) Describe the overall effect of the mapping:
The mapping resulted in the square in the z-plane being rotated 45 degrees counter-clockwise. The length of the side of the square was increased by a factor of $\sqrt{2}$ and the area of the square was doubled.

Question 7

Find the image of the unit square $0\leq x\leq 1, 0\leq y\leq 1,$ under the mapping $W=Z^{2}$

WIKI%20HW%20FUNCTIONS.png

$W=Z^{2} -> W=(x+iy)^{2} -> u+iv = x^{2}+2ixy-y^{2}$
$u = x^{2}-y^{2}$
$y=2yx$

Side 1: $0\leq x\leq 1, y=0$ -> {u=x^{2}, v=0}
Positive real axis up to 1.

Side 2: $x=1, 0\leq y\leq 1,$ -> {u=1-y^{2},v=2y}
$y=v/2 ->u=1-(v^{2}/4)$
Parabola

Side 3: $0\leq x\leq 1, y=1$ -> {u=x^{2}-1, v=2x}
$x=v/2, u=(v^{2}/4)-1$
Parabola

Side 4: $x=0, 0\leq y\leq 1,$ -> {u=-y^{2}, v=0}
Negative real axis up to 1

Question 8

Show that the Joukowski transformation maps the circle $|z|=r$ onto the ellipse $\frac{u^2}{(\frac{1}{2}(r+\frac{1}{r}))^2}+\frac{v^2}{(\frac{1}{2}(r-\frac{1}{r}))^2}�788�$.

(34)
\begin{align} j(z)=\frac{1}{2}(z+\frac{1}{2})\\ z=x+iy\\ j(z)=\frac{1}{2}(x+iy+\frac{1}{2(x+iy)})\\ j(z)=\frac{1}{2}x+\frac{1}{2}iy +\frac{x-iy}{2(x+iy)(x-iy)}\\ j(z)=\frac{1}{2}x+\frac{1}{2}iy +\frac{x-iy}{2(x^2+y^2)}\\ j(z)=\frac{1}{2}x+\frac{x}{2(x^2+y^2)} +\frac{1}{2}i(y-\frac{y}{(x^2+y^2)})\\ \end{align}

This gives you:

(35)
\begin{align} u=\frac{1}{2}x+\frac{x}{2(x^2+y^2)}\\ v=\frac{1}{2}y+\frac{y}{2(x^2+y^2)}\\ \end{align}

Use Euler's identity:

(36)
\begin{align} u=\frac{1}{2} r cos(\theta)+\frac{cos(\theta)}{2r}\\ u=cos(\frac{1}{2}r+\frac{1}{2r})\\ v=\frac{1}{2} r sin(\theta)-\frac{sin(\theta)}{2r}\\ \end{align}

Rearrange it to solve for cosine and sine.

(37)
\begin{align} cos(\theta)=\frac{u}{\frac{1}{2}(r+\frac{1}{r})}\\ sin(\theta)=\frac{v}{\frac{1}{2}(r-\frac{1}{r})}\\ \end{align}

Applying this to the identity $cos^2(\theta) +sin^2(\theta)=1$ gives you:

(38)
\begin{align} \frac{u^2}{(\frac{1}{2}(r+\frac{1}{r}))^2}+\frac{v^2}{(\frac{1}{2}(r-\frac{1}{r}))^2}\\ \end{align}

Here is a graph of it:

Joukowski%20Airfoil%20Geogebra.png
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