(Another) Calculus Question

Sully

Hey all,

Quick question:-

What is the solution for evaluating (3+4i)^10 or cubic root of (3+4i).

Its been a few years since I did it, but I always thought it was easy. But I cant find it in leaving cert text books, and my notes on Complex Numbers dont mention it. I know about expressing in polar form and expotential when they are in the Cartesian form but moving from such an example into cartesian I cant recall.

Any solutions? Cheers.

TX123

First you have to get (3+4i) into polar form. on x axis have 3 and y axis have 4. find hypotenuse(hyp). find the tan of the angle.
The use then write the sum in the form of :
Hypotenuse(Cos(angle) -iSin (angle)

=hyp(cis(angle)^10)

Put the hyp to the power of 10 and multiply the angle by 10.

Find the angle by taking 360' away as many times as you can before you get less that 360 as your angle.

E.G
2^6cis(-540+360+360)

= 64cis180'
=64(cos 180 - i sin 180)
=64(-1+0i)= -64

Sully

Whats "cis"? Im assuming the first few lines are working out a polar form, or else its very similar!

What exactly is this called?

Fallen Seraph

I *think* I've come across the cis notation before cis(theta) corresponds to cos(theta)+iSin(theta). It makes sense in this context, anyway.

TX123

Fallen Seraph wrote:

I *think* I've come across the cis notation before cis(theta) corresponds to cos(theta)+iSin(theta). It makes sense in this context, anyway.

ye cis is exactly that. just a shortened version. sorry for the confusion.

aequinoctium

notation: so handy yet so confusing

Farouk.Bulsara

Sully wrote:

Hey all,

Quick question:-

What is the solution for evaluating (3+4i)^10 or cubic root of (3+4i).

Its been a few years since I did it, but I always thought it was easy. But I cant find it in leaving cert text books, and my notes on Complex Numbers dont mention it. I know about expressing in polar form and expotential when they are in the Cartesian form but moving from such an example into cartesian I cant recall.

Any solutions? Cheers.

Hi Sully,

In order to calculate powers (or roots) of complex numbers, you need to do two things:

1. Convert the number from Cartesian form (a + ib) into Polar form ( r [ cos theta + i sin theta ]

2. Apply De Moivre's Theorem to the result. De Moivre's Theorem states that r [ cos theta + i sin theta] ^ n is equal to r ^ n [ cos n*theta + i sin n*theta ]


Hope this helps.

Fred

Michael Collins

aequinoctium wrote:

notation: so handy yet so confusing

And even easier to write is hyp*e^[i*(theata)], which also equals "cos(theata)+i*sin(theata)". Where e is natural log base and i is the square root of -1.

Maybe this will make your calculations even more straightforward...

Sully

Farouk.Bulsara wrote:

Hi Sully,

In order to calculate powers (or roots) of complex numbers, you need to do two things:

1. Convert the number from Cartesian form (a + ib) into Polar form ( r [ cos theta + i sin theta ]

2. Apply De Moivre's Theorem to the result. De Moivre's Theorem states that r [ cos theta + i sin theta] ^ n is equal to r ^ n [ cos n*theta + i sin n*theta ]


Hope this helps.

Fred

Ah, thanks a million. I knew it was straight forward but for the life of me I couldn't remember. Thanks.