## Monthly Archives: July 2015

### Complex Numbers for you

If $iz^{3}+z^{2}-z+i=0$, then $|z|$ equals

(a) 4

(b) 3

(c) 2

(d) 1.

Solution.

We can write the given equation as $z^{3}+\frac{1}{i}z^{2}-\frac{1}{i}z+1=0$, or $z^{3}-iz^{2}+iz-i^{2}=0$ $\Longrightarrow z^{2}(z-i)+i(z-i)=0$ $\Longrightarrow (z^{2}+i)(z-i)=0 \Longrightarrow z^{2}=-i, z=i$ $\Longrightarrow |z|^{2}=|-i|$ and $|z|=|i|$ $\Longrightarrow |z|^{2}=1$ and $|z|=1$ $\Longrightarrow |z|=1$

More later,

Nalin Pithwa

### River Crossing 1 — Farm Produce

Alcuin of Northumbria, aka Flaccus Albinus Alcuinus or Ealhwine, was a scholar, a clergyman and a poet. He lived in the eighth century and rose to be a leading figure at the court of the emperor Charlemagne. He included this puzzle in a letter to the emperor,  as an example of ‘subtlety in Arithmetick for your enjoyment’. It still has mathematical significance, as I will eventually explain. It goes like this.

A farmer is taking a wolf, a goat, and a basket of cabbages to market,, and he comes to a river where there is a small boat. He can fit only one item out of the three into the boat with him at any time. He can’t leave the wolf with the goat, or the goat with the cabbages, for reasons that should be obvious. Fortunately, the wolf detests cabbages. How does the farmer transport all three items across the river?

Have fun!

Nalin Pithwa

### More complex stuff

Problem.

If $z_{1}, z_{2}, \ldots , z_{n}$ lie on the unit circle $|z|=2$, then value of $E=|z_{1}+z_{2}+\ldots+z_{n}|-4|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots+\frac{1}{z_{n}}|$ is

(a) 0

(b) n

(c) -n

(d) none of these.

Solution.

As $z_{1},z_{2},\ldots, z_{n}$ lie on the circle $|z|=2$, $|z_{i}|=2 \Longrightarrow |z_{i}|^{2}=4 \Longrightarrow z_{i}\overline{z_{i}}=4$ for $i=1,2,3, \ldots, n$

Thus, $\frac{1}{z_{i}}=\frac{\overline{z_{i}}}{4}$ for $i=1, 2, 3, \ldots, n$

Hence, $E=|z_{1}+z_{2}+\ldots+z_{n}|-4|\frac{\overline{z_{1}}}{4}+\frac{\overline{z_{2}}}{4}+\ldots+\frac{\overline{z_{n}}}{4}|$, which in turn equals $|z_{1}+z_{2}+\ldots+z_{n}|-|\overline{z_{1}}+\overline{z_{2}}+\ldots+\overline{z_{3}}|$, that is, $|z_{1}+z_{2}+\ldots+z_{n}|-|\overline{z_{1}+z_{2}+\ldots+z_{n}}|=0$.

(since $|z|=|\overline{z}|$).

More later,

Nalin Pithwa

### Rabbits in the Hat

The Great Whodunni, a stage magician, placed his top hat on the table.

‘In this hat are two rabbits,’ he announced. ‘Each of them is either black of white, with equal probability. I am now going to convince you, with the aid of my lovely assistant Grumpelina, that I can deduce their colours without looking inside the hat!’

He turned to his assistant, and extracted a black rabbit from her costume. ‘Please place this rabbit in the hat.’ She did.

Whodunnii now turned to the audience. ‘Before Grumpelina added the third rabbit, there were four equally likely combinations of rabbits. ‘ He wrote a list on a small blackboard:BB, BW, WB and WW. ‘Each combination is equally likely — the probability is 1/4.

But, then I added a black rabbit. So, the possibilities are BBB, BWB, BBW and BWW — again, each with probability 1/4.

‘Suppose —- I won’t do it, this is hypothetical — suppose I were to pull a rabbit from the hat. What is the probability that it is black? If the rabbits are BBB, that probability is 1. If BWB or BBW, it is 2/3. If BWW, it is 1/3. So the overall probability of pulling out a black rabbit is $\frac{1}{4} \times 1 + \frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{1}{3}$

which is exactly 2/3.

‘But. If there are three rabbits in a hat, of which exactly r are black and the rest white, the probability of extracting a black rabbit is r/3. Therefore, $r=2$, so there are two black rabbits in the hat.’ He reached into the hat and pulled out a black rabbit. ‘Since I added this black rabbit, the original pair must have been one black and one white!’

The Great Whodunni bowed to tumultous applause. Then, he pulled out two rabbits fwrom the hat — one pale lilac and the other shocking pink.

It seems evident that you can’t deduce the contents of a hat without finding out what’s inside. Adding the extra rabbit and then removing it again (was it the same black rabbit? Do we care?) is a clever piece of misdirection. But, why is the calculation wrong?

More later,

Nalin Pithwa

### Shaggy Cat Story

No cat has eight tails,

One cat has one tail.

Adding: one cat has nine tails.

🙂 🙂 🙂

More later,

Nalin Pithwa

### De Moivre’s Theorem application

Question:

If $f_{r}(\alpha)=(\cos{\frac{\alpha}{r^{2}}}+i\sin{\frac{\alpha}{r^{2}}}) \times (\cos{\frac{2\alpha}{r^{2}}}+i\sin{\frac{2\alpha}{r^{2}}}) \ldots (\cos{\frac{\alpha}{r}}+i\sin{\frac{\alpha}{r}})$, then $\lim_{n \rightarrow \infty}f_{n}{\pi}$ equals

(a) -1

(b) 1

(c) -i

(d) i

Solution.

Using De Moivre’s theorem, $f_{r}{\alpha}=e^{i\frac{\alpha}{r^{2}}}e^{i\frac{2\alpha}{r^{2}}}\ldots e^{i\frac{\alpha}{r}}$

which in turn equals $e^{(i \frac{\alpha}{r^{2}})(1+2+\ldots+r)}=e^{(i\frac{\alpha}{r^{2}})(\frac{r(r+1)}{2})}=e^{i(\frac{\alpha}{2})(1+\frac{1}{r})}$

Hence, $\lim_{n \rightarrow \infty}f_{n}(\pi)=\lim_{n \rightarrow \infty}e^{(i)(\frac{\pi}{2})(1+\frac{1}{n})}=e^{i(\frac{\pi}{2})}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i$.

More complex stuff to be continued in next blog (pun intended) 🙂

Nalin Pithwa

### A complex equation

Find the number of solutions of the equation $z^{3}+\overline{z}=0$.

Solution.

Given that $z^{3}+\overline{z}=0$. Hence, $z^{3}=-z$. $|z|^{3} =|-\overline{z}| \Longrightarrow |z|^{3}=|z|$.Hence, we get $|z|(|z|-1)(|z|+1)=0 \Longrightarrow |z|=0, |z|=1$ (since $|z|+1>0$)

If $|z|=1$, we get $|z|^{2}=1 \Longrightarrow z.\overline{z}=1$.

Thus, $z^{3}+\overline{z}=0 \Longrightarrow z^{3}+1/z=0$

Thus, $z^{4}+1=0 \Longrightarrow z^{4}=\cos{\pi}+i\sin{\pi}$, that is, $z=\cos{\frac{2k+1}{4}}\pi+i\sin{\frac{2k+1}{4}}\pi$ for $k=0,1,2,3$. Therefore, the given equation has five solutions.

### Shaggy Dog Story

Brave Sir Lunchalot was travelling through foreign parts. Suddenly, there was a flash of lightning and a deafening crack of thunder, and the rain started bucketing down. Fearing rust, he headed for the nearest shelter, Duke Ethelfred’s castle. He arrived to find the Duke’s wife, Lady Gingerbere weeping piteously.

Sir Lunchalot liked attractive young ladies, and for a brief moment he noticed a distinct glint through Gingerbere’s tears. Ethelfred was very old and frail, he observed…Only one thing, he vowed would deter him from a secret tryst with the Lady — the one thing in all the world that he could not stand.

Puns.

Having greeted the Duke, Lunchalot enquired why Gingerbere was so sad.

“It is my uncle Elpus,” she explained. “He died yesterday.”

“Permit me to offer my sincerest condolences,” said Lunchalot.

“That is not why I weep so…so piteously, sir knight,” replied Gingerbere. “My cousins Gord, Evan and Liddell are unable to fulfill the terms of uncle’s will.”

“Why ever not?”

“It seems that Lord Elpus invested the entire family fortune in a rare breed of giant-riding dogs. He owned 17 of them.”

Lunchalot had never heard of a riding-dog, but he did not wish to display his ignorance in front of such a lithesome lady. But, this fear, it appeared, could be set to rest, for she said, “Although I have heard much of these animals, I  myself have never set eyes on one.”

“They are no fit sight for a fair lady,” said Ethelfred firmly.

“And, the terms of the will —?” Lunchalot asked, to divert the direction of the conversation.

“Ah, Lord Elpus left everything to the three sons. He decreed that Gord should receive half the dogs, Evan one third, and Liddell one ninth.”

“Mmm. Could be messy.”

“No dog is to be subdivided, good knight.”

Lunchalot stiffened at the phrase good knight, but decided it had been uttered innocently and was not a pathetic attempt at humour.

“Well, —- : Lunchalot began.

“Pah, ’tis a puzzle as ancient as yonder hills!” said Ethelfred scathingly. “All you have to do is take one of your riding dogs over to the castle. Then, there are 18 of the damn things!”

“Yes, my husband, I understand the numerology, but —”

“So, the first son gets half that, which is 9; the second gets one third which is 6; the third son gets one ninth, which is 2. That makes 17 altogether, and our own dog can be taken back here!”

“Yes, my husband, but we have no one here who is manly enough to ride such a dog.”

Sir Lunchalot seized his opportunity. “Sire, I will ride your dog!” The look of admiration in Gingerbere’s eye showed him how shrewd his gallant gesture had been.

“Very well,” said Ethelfred.”I will summon my houndsman and he will bring the animal to the courtyard. Where we shall meet them.”

They waited in an archway as the rain continued to fall.

When the dog was led into the courtyard, Lunchalot’s jaw dropped so far that it was a good job he had his helmet on. The animal was twice the size of an elephant, with thick striped fur, claws like broadswords, blazing red eyes the size of Lunchalot’s shield, huge floppy ears dangling to the ground, and a tail like a pig’s — only with more twists and covered in sharp spines. Rain cascaded off its coat in waterfalls. The smell was indescribable.

Perched improbably on its back was  a saddle.

Gingerbere seemed even more shocked than he by the sight of this terrible monstrosity. However, Sir Lucnhalot was undaunted. Nothing could daunt his confidence. Nothing could prevent a secret tryst with the lady, once he returned astride the giant hound, the will executed in full. Except…

Well, as it happened, Sir Lunchalot did not ride the monstrous dog to Lord Elpus’s castle, and for all he knows the will has still not been executed. Instead, he leaped on his horse and rode off angrily into the stormy darkness, mortally offended, leaving Gingerbere to suffer the pangs of unrequited lust.

It wasn’t Ethelfred’s’ dodgy arithmetic — it was what the Lady had said to her husband in a stage whisper.

What did she say?

🙂 🙂 🙂

More later,

Nalin Pithwa

### More trigonometry practice

Problem.

Let ABC be a triangle and $h_{a}$ be the altitude through A. Prove that $(b+c)^{2} \geq a^{2}+4(h_{a})^{2}$.

(As usual, a, b, c denote the sides BC, CA, AB respectively.)

Proof.

The given inequality is equivalent to $(b+c)^{2}-a^{2} \geq 4(h_{a})^{2}=\frac{16\triangle^{2}}{a^{2}}$.

where $\triangle$ is the area of triangle ABC. Using the identity $16\triangle^{2}=[(b+c)^{2}-a^{2}][a^{2}-(b-c)^{2}]$, we see that the inequality to be proved is $a^{2}-(b-c)^{2} \leq a^{2}$ (here we use $a), which is true. Observe that equality holds iff $b=c$. QED.

More later,

Nalin Pithwa

A cute problem in Trigonometry or pure plane geometry Mathematics Hothouse

Problem.

In a triangle ABC, $latex angle A$ is twice $latex angle B$. Show that $latex a^{2}=b(b+c)$. (In fact, the converse is also true. Prove it!).

Proof.

Method I.

You can use plane geometry also. This is left to you as an exercise.

Method II.

You can use trigonometry also. We may use the sine rule for a triangle to dispose of both the implications simultaneously.

$latex A=2B Longleftrightarrow A-B=B Longleftrightarrow sin{(A-B)}=sin{B} Longleftrightarrow sin{(A-B)}sin{(A+B)}=sin{B}sin{C} Longleftrightarrow sin^{2}{A}-sin^{2}{B}=sin{B}sin{C} Longleftrightarrow (2Rsin{A})^{2}-(2Rsin{B})^{2}=(2Rsin{B})(2Rsin{C}) Longleftrightarrow a^{2}-b^{2}=bc Longleftrightarrow a^{2}=b(b+c)$

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