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i i
By Mark Woodward (P1738) on Friday, December 17, 1999 - 11:35 am:

Please could somebody tell me if the following workings are correct, or if I am using properties that don't hold for complex numbers.

eix = cosx + isinx
If x=p/2:
eip/2 = i
Raising both sides to power i:
(eip/2)i = ii
By law of indices:
e-p/2 = ii
This is a bit strange since the LHS represents a real number.

Is this right or have I gone completely wrong?!

By Dan Goodman (Dfmg2) on Friday, December 17, 1999 - 12:33 pm:

This is right, there's only one other thing to say, and that is that this is not the only value for ii. eip/2+2npi=i for all integers n. Therefore ii=e-p/2-2np for all integers n. In general, when you do things like raising complex numbers to the power of other complex numbers, you get an infinite number of values. You might like to try to think about how you work out (a+ib)c+id and why this leads to an infinite number of values. (Hint: take logs). If you can't do this, post again and I'll f
ill in the details.

By Mark Woodward (P1738) on Tuesday, January 11, 2000 - 09:52 am:

I've tried to work out (a+ib)(c+id), but cannot get a completely real solution - I can only get parts of it to be real. Please help me!

By Dan Goodman (Dfmg2) on Tuesday, January 11, 2000 - 08:10 pm:

Oops, I think I may have misled you, you don't always get a real solution, ii is special in that respect. In fact, if you have the form for (a+ib)c+id you could probably find the condition for it to be real reasonably easily. Here's another hint, let z=a+ib and w=c+id, then we want to work out zw, we know that zw=ewlog(z) for real w,z and this is also true for complex w,z. Now we need to work out what log(z) is, well, if you write z as reiq then clearly log(z)=log(r)+iq. In other words, log(z)=log(abs(z))+iArg(z) where Arg(z) is the angle of z. Do you want to try to finish this off or shall I finish it up?

By Mark Woodward (P1738) on Thursday, January 13, 2000 - 01:55 pm:

I hope this is what you were thinking of.
wlog(z)=(c+id)log(z)
=(c+id)(log(abs(z))+iArg(z))
=(c+id)(log(a2+b2)1/2+iTan-1(b/a))
So therefore (a+ib)c+id=
e(c+id)(log(a2+b2)1/2+iArctan(b/a))
Aaaagh!
This gives the special case when a=c=0 and b=d=1 of e-p/2

By Dan Goodman (Dfmg2) on Thursday, January 13, 2000 - 06:22 pm:

That's right, but you didn't put in the bit about multiple values of log(z). When you find the angle Arctan(b/a) there are two problems. First of all, this only gives the correct angle in two quadrants as (-b)/(-a)=b/a, which is why we define the function Arg(z) to be the correct angle in all 4 quadrants. Secondly, there are an infinite number of possible values for the angle, Arg(z)+2np for any n. If you now put this into the equation you will find more special cases where it is real. Another slight simplification is log(sqrt(a2+b2))=(1/2)log(a2+b 2). So, expanding out we get: (a+ib)c+id=exp((c+id)((1/2)log(a2+b2)+i(Arg(a+ib)+2n p)). If we put the imaginary part of this expression equal to 0, we get

0=exp(i((d/2)log(a2+b2)+c(Arg(a+ib)+2n p)))
0=sin((d/2)log(a2+b2)+c(Arg(a+ib)+2n p))

So, when this holds, it will be real. So one condition we could write for the thing to be real is. There are two integers, n and m such that:

(d/2)log(a2+b2)+c(Arg(a+ib)+2np)=mp

If you're still interested, see if you can simplify this condition into something nicer. By the way, another good special case when it is real is when b=d=0, a>=0.

It may seem strange that there isn't a unique value of (a+ib)c+id, and it is. A function like log(z) which has this property can be split into what are called branches, each different value of n is a different branch. What you can do is take what is called the "principal branch", this is denoted Log(z) and is the one where the angle is in the interval [-p,p). This specifies a unique value of (a+ib)c+id.