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Welcome to NRICH.

What is e^z?

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By Anonymous on Friday, September 15, 2000 - 07:27 pm:

Hi
If z is a complex number, what would e^z mean?
Thanks

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By Michael Doré (P904) on Friday, September 15, 2000 - 09:33 pm:

Well if z = a + bi where a,b are real then we have:

e^z = e^a × e^bi = e^a (cos b + i sin b)

because e^bi = cos b + i sin b

by Euler's formula. My favourite way of proving Euler's formula is using second order differential equations.

Let f(x) = e^ax

Easy to see:

f''(x) = a²e^ax

So:

f''(x) + f(x) = e^ax [1+a²]

Now let a = i, so that f(x) = e^ix and a² = -1. Then:

f''(x) + f(x) = 0.

This is the second order simple harmonic motion differential equation (which is used in analysing springs, for instance). The general solution can be easily shown as:

f(x) = A cos x + B sin x

So:

e^ix = A cos x + B sin x

where A,B are constant.

Setting x = 0:

e⁰ = A

So A = 1. Differentiate:

ie^ix = -A sin x + B cos x

Set x = 0:

i = B. So:

e^ix = cos x + i sin x

as required.

Yours,

Michael

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By Anonymous on Saturday, September 16, 2000 - 02:28 pm:

What would the transfomation of Z going to e^Z do to the line of Re(z) = 2 and why?
Plus where on an Argand diagram would e^z=0 ?

Thanks for all your help.

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By Pras Pathmanathan (Pp233) on Monday, September 18, 2000 - 09:24 pm:

Hi Anon.

The line Re(z)=2 is given by z=2+iy for y real. So e^z is:

e^z = e^2+iy = e²e^iy = e²(cos y + isin y)

Since e² is greater than 0, this is in modulus-argument form, modulus e² and argument y. y can be any real number, so certainly any possible argument is possible, so Re(z) maps to the set of points of modulus e², ie the circle of radius e² (and centre 0).

As for your second question, there isn't any complex number z such that e^z = 0. For suppose there was, and write z as z=x+iy, where x and y are real. Then:

0 = e^z = e^x(cosy+isiny) = e^xcosy + ie^xsiny

Equating real and imaginary parts gives

e^xcosy = 0
e^xsiny = 0 where x,y are real

cos y and sin y are never both zero, so we must have e^x=0. But x is real, and we know that e^x > 0 for real x, so no such z is possible.

Pras

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By Tim Garton (P2843) on Tuesday, September 19, 2000 - 07:55 pm:

Thank you

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