Your personal integral

Let's use a little numerology to generate your own personal integral
$$\int_{\gamma} f(z) \, dz.$$
The idea is to compute the integral in two ways and show the results are the same. Thus,

• Compute the integral by translating it to an ordinary integral of a function mapping $\mathbb R \to \mathbb C$, and
• Compute the integral using a complex anti-derivative.

To generate your personal integral, we'll first compute two integers $m$ and $n$ as follows:

• $m$ is the sum of the digits of the sum of the digits of your Student ID number.
• $n$ is the the name number associated with just your first name, as computed in the standard but crazy numerological way.

We then define

• $a=n\mod 3+2$
• $b=n\mod 4+1$
• $c=m\mod 3+1$
• $p=m$

Finally, your personal path is the line segment from the origin to the point $a+bi$ and your personal function is
$$f(z) = z^p + cz.$$

OK - here's my attempt. (Of course, the first thing I did was give a "like" to the original question, because that's the polite thing to do.)

First, my student ID number (which you should not post in your answer; mine is just made up after all) is $987654321$. The sum of those digits is $45$ and the sum of $4$ and $5$ is $9$. So, for me, $m=9$.

Next, my first name is Audrey and letters of my name occur at the positions 1, 21, 4, 18, 5, and 25 of the alphabet. The sum of those numbers gives me $n=74$.

Using all that, I get 4, 3, 1, 74

• $a=n\mod 3+2 = 4$
• $b=n\mod 4+1 = 3$
• $c=m\mod 3+1 = 1$
• $p=m = 74$

Putting that all together, I get
$$\int_{\gamma} (z^{74} + z) \, dz,$$
where $\gamma$ is the line segment from zero to $4+3i$.

When I evaluate that, I get
$$\frac{1}{75}(4+3i)^{75} + \frac{1}{2} (4+3i)^2.$$
I guess maybe you should show more work on this part, though.

My numbers are $m=6$ and $n=45$.

So my values for $a$, $b$, $c$ and $p$ are
$$
\begin{align}
a&=45\mod3+2=2\\
b&=45\mod4+1=2\\
c&=6 \ \ \mod3+1=1\\
p&=6
\end{align}
$$
My path is the line segment from the origin to the point $2+2i$ and my function is
$$
f(z)=z^6+z.
$$
My integral is therefore
$$
\int_\gamma(z^6+z) \ dz,
$$
where $\gamma$ is the line segment from zero to $2+2i$.

Computing the integral using a complex anti-derivative gives
$$
\frac{1}{7}(2+2i)^7+\frac{1}{2}(2+2i)^2=\frac{1024}{7}-\frac{996}{7}i.
$$
We could also translate this complex integral to an ordinary integral of function mapping $\mathbb{R}\rightarrow\mathbb{C}$ by making use of the fact that
$$
\int_\gamma f(z) \ dz=\int_a^bf(\gamma(t)) \ \gamma'(t) \ dt,
$$
where $\gamma(t)=(2+2i)t$ and $\gamma'(t)=2+2i$ so $a=0$ and $b=1$.

Thus
$$
\int_\gamma(z^6+z) \ dz=\int_0^1\big(((2+2i)t)^6+(2+2i)t\big)(2+2i) \ dt.
$$
Evaluating this integral gives
$$
(2+2i)\bigg(\frac{(2+2i)^6}{7}+\frac{(2+2i)}{2}\bigg)=\frac{1024}{7}-\frac{996}{7}i.
$$
Note that both methods of computing the integral give the same answer.

My numbers are $m = 9$ and $n = 2$. This gives

$a = 4$
$b = 3$
$c = 1$
$p = 9$

My path from the origin to $4+3i$ is $\gamma(t) = (4+3i)t$ with $0\leq t \leq 1$. My function is $f(z) = z^9 + z$.
The integral with respect to $z$ is shown below, it can be evaluated with respect to $t$ as $$\int_{\gamma} f(z) dz = (4+3i) \int_{0}^{1} (4+3i)^9 t^9 + (4+3i) t dt $$

We take the antiderivative and evaluate at $1$ and subtract it evaluated at $0$ to get $$\int_{\gamma} f(z) dz = (4+3i) \int_{0}^{1} (4+3i)^9 t^9 + (4+3i)^t dt = \frac{(4+3i)^{10}}{10} + \frac{(4+3i)^2}{2}$$

This can be simplified to $$ \frac{4826661}{5} + \frac{738552 i}{5}$$

Now we compute the integral using a complex anti-derivative of $f(z) = z^9 + z$. The anti-derivative $F(z) = \frac{z^{10}}{10} + \frac{z^2}{2}$ is valid antiderivative and we compute the integral as $F(4+3i) - F(0)$ = $F(4+3i)$.

This gives us
$$\int_{\gamma} f(z) dz = F(4+3i) = \frac{4826661}{5} + \frac{738552i}{5}$$ which is equivalent to the other method of finding the integral.

@lszabo This solution addresses the first part of the prompt well. However, you still need to compute the integral by using the complex anti-derivative then plugging in your value of $z$. This is to show that whether we compute the integral in the complex plane (inputs complex numbers and outputs complex numbers) or we first translate it to an ordinary integral (inputs real numbers and outputs complex numbers), we should get the same result.

My numbers are $n=34$ and $m=5$. This means
\begin{align*}
a=3&&b=3&&c=3&&p=5
\end{align*}
Therefore, $f(z)=z^5+3z$ and $\gamma$ is the line segment from the origin to $3+3i$ and $\gamma(t)=(3+3i)t$. Note $\gamma'(t)=3+3i.$. For neatness sake, let $a=3+3i$.

I will start by integrating using complex anti-derivatives.
\begin{align*}
\int_\gamma f(z)&=\int_{o}^{a}z^5+3z\ dz\\
&=\frac{1}{6}\, z^6+\frac{3}{2}\,z^2\Big|_{0}^{a}\\
&=\frac{1}{6}\, a^6+\frac{3}{2}\,a^2-0\\
&=\frac{1}{6}(-5832i)+\frac{3}{2}\,(18i)\\
&=-945i
\end{align*}
Now I will use the formula $$\int_{0}^{a}f(z)=\int_{t_o}^{t_f}f(z)\big(\gamma(t)\big)\gamma'(t)dt$$
This means
\begin{align*}
\int_{\gamma}f(z)&=\int_{0}^{1}f(at)a\ dt\\
&=a\int_{0}^{1}(at)^5+3at\\
&=a\bigg(\frac{1}{6}a^5t^5+\frac{3}{2}a\bigg|_{0}^{a}\bigg)\\
&=a\big(a^5+3a-0)\\
&=\frac{1}{6}a^6+\frac{3}{2}a^2\\
&=\frac{1}{6}(-5832i)+\frac{3}{2}(18i)\\
&=-945i
\end{align*}

My personal integral will be generated using m=5 and n=53. So,
$a=53$ mod $3+2=4,$
$b=53$ mod $4+1=2,$
$c=5$ mod $3+1=3,$ and
$p=m=5.$
So my path is the line segment from the origin to the point 4+2i and my personal integral is $$\int_{0}^{4+2i} (z^5+3z)dz$$
Evaluating the integral yields $$\frac{1}{6}(4+2i)^6+\frac{3}{2}(4+2i)^2=-1230+493.333i$$
We can also take advantage of the fact that $$\int_{\gamma}f(z)dz=\int_{b}^{a}f(\gamma(t))\gamma'(t)dt$$ where $\gamma(t)=(2+i)t$ and $\gamma'(t)=2+i$ and $0<t<2$, so $a=0$ and $b=2$.
So, $$\int_{0}^{2}(((2+i)t)^5+(3(2+i)t))*(2+i)dt$$
and $$(\frac{1}{6(2+i)}((2+i)2)^6+\frac{3}{2(2+i)}((2+i)2)^2)*(2+i)=\frac{1}{6}(4+2i)^6+\frac{3}{2}(4+2i)^2=-1230+493.333i$$

I have $m=4$, $n=48$. So: $$a=48\mod 3 +2= 2 \\b = 48\mod 4+1=1 \\c = 4 \mod 4 +1=2 \\p=4$$

Therefore,
$$ \int_\gamma (z^4+2z)dz$$
where $\gamma$ is the line segment from the origin to $2+i$.

Solving the integral,
$$ \frac{1}{5}z^5+z^2|^{(2+i)}_0
\\ = \frac{1}{5}(2+i)^5+(2+i)^2
\\ = -4.6+12.2i $$

Solving another way, $\gamma(t)=(2+i)t$ and $\gamma'(t)=(2+i)$ which will be evaluated from $0\leq t\leq1$.

$$\int_0^1 (((2+i)t)^4+2((2+i)t))*(2+i)dt
\\= \frac{(2+i)^5}{5}+(2+i)^2
\\=-4.6+12.2i$$

My personal integral will be generated using the seed values $m=10$ and $n=110$.
Then, $a=110\mod 3+ 2=4$,
$b=110\mod 4 + 1 = 3$,
$c=10 \mod 3 + 1 = 2$,
$p=m=10$.

Accordingly, $f(z)=z^{10}+2z$ and $\gamma$ is defined as the line segment from the origin to $4+3i$.

Then, my personal integral is $$ \int_{\gamma} (z^{10}+2z) dz. $$

Using the method of complex anti-derivative, we calculate \begin{align} \int_{\gamma} (z^{10}+2z) dz&=\int_{0}^{4+3i} (z^{10}+2z)dz \\
&= \biggr(\frac{z^{11}}{11} + z^2 \biggr)\biggr\rvert^{4+3i}_0 \\
&=\frac{(4+3i)^{11}}{11}+(4+3i)^2. \end{align}

We now seek to confirm our result by calculating this integral after translating it to an ordinary integral by way of parameterization of $\gamma$ in terms of $t$. Let $\gamma(t)=(4+3i)t$ where $ 0\leq t \leq 1, \Rightarrow \gamma '(t)=4+3i$. Then we can use the relationship $\int_\gamma f(z) dz = \int_{a}^b f(\gamma(t))\gamma'(t) dt$ to transform our integral into $$\int_0^{1} [((4+3i)t)^{10}+2((4+3i)t)]*(4+3i) dt \\
=\int_0^1 (4+3i)^{11}t^{10}+2*(4+3i)^2t dt \\
=\frac {(4+3i)^{11}t^{11}}{11} + (4+3i)^2t^2 \biggr\rvert_0^1 \\
=\frac{(4+3i)^{11}}{11}+(4+3i)^2$$

As we can see that this solution does agree with the first method, we conclude that with either method we may calculate the value of the personal integral to be $$\frac{(4+3i)^{11}}{11}+(4+3i)^2=\frac{34182273}{11}+34867821i.$$

Ok so my personal numbers are the following: \[\begin{aligned}
&c=1 \quad a=4\quad b=1 \quad p=6\end{aligned}\]
Thus my personal integral is \(\int_{\gamma} (z^{6} + z) \, dz\) where \(\gamma=4+i\). Thus the anti-derivative is \(\frac{1}{7}(4+i)^7+\frac{1}{2}(4+i)^2\) which solves to \(\frac{-5711}{14}+\frac{20075i}{7}\) Evaluating by integration requires setting \(\gamma(t)=(4+i)t\) and \(\gamma'(t)(4+i)\) creating \[\int_0^1 [((4+i)t)^6+((4+i)t)](4+i)\, dz\] which equals \[(4+i)(\frac{(4+i)^6}{7}+\frac{4+i}{2})\quad = \frac{-5711}{14}+\frac{20075i}{7}\] Which is the same!

Using your method I got:

$\begin{array}{lrcl}
\bullet&a&=&2\\
\bullet&b&=&1\\
\bullet&c&=&1\\
\bullet&p&=&9\\
\end{array}$

Which makes my personal integral:
$$\int_\gamma z^9+z$$

Where $\gamma$ is the path from the origin to $2+i$.
Using the complex anti-derivative I get:
$$\int_0^{2+i}z^9+z=\frac{1}{10}(2+i)^{10}+\frac{1}{2}(2+i)^2$$
But I'm not sure if this integral is path independent. Since this isn't a terribly hard integral, I think I'll make my path exactly as it is defined and use:
$$\int_\gamma f(z)dz=\int_a^b f\left(\gamma (t)\right)\gamma '(t)dt$$
If I use $\gamma (t)=(2+i)t$, I can use bounds $a=0$ and $b=1$. I also get $\gamma '(t)=2+i$. Solving uses a simple $u$-substitution.
Starting with:
$$\int_\gamma z^9+z=(2+i) \int_0^1 \left(\left((2+i)t\right)^9+(2+i)t\right) dt$$
I choose:
$$\begin{array}{rcl}
u&=&(2+i)t\\
du&=&(2+i)dt
\end{array}$$
So,after changing my bounds,I now have:
$$\int_0^{2+i}(u^9+u)du=\frac{(2+i)^{10}}{10}+\frac{(2+i)^2}{2}$$
Which is the same as before!