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Dear Prof Vijilant

Can we compute exponentiation of two complex numbers?

Example - 1+2i and 2+3i

I.e 1+2i raised to power of 2+3i

Thanks

Prashant

Hello Prashant

I put the expression (1+2i)^(2+3i) into my advanced TI92 calculator and it came up with the answer -.015133 - 0.179867i. But could I have done that with an ordinary scientific calculator? The answer of course is yes. First, the laws of exponentiation. a^x * a^y = a^(x +y). (x^m)^n = x^(m+n) will be needed as well as the polar and exponential forms of a complex number and De Moivres theorem.

On the argand diagram x + iy is represented by the point (a,b), but if polar coordinates are used it becomes R(cos(t) + i*sin(t)) where R = (x^2 + y^2)^(1/2) and t = invtan(y/x). using the Taylor series for sin and cos gives the exponential form R*e^(it). Theta is the usual angle, but I can't type it so have used t.

Moivres theorem is (cos(t) + i*sin(t))^n = cos(nt) + i*sin(nt) which is easily proved using the exponential form.

We first need the polar form of 1 + 2i. It is 5^(1/2)*(cos(t) + i*sin(t)) where t = invtan(2).Then

(1+2i)^(2+3i) = ((1+2i)^2)*(1+2i)^(3i). Using the polar form, and using De Moivre we have

5(cos(2t)+i*sin(2t)*(5^(1/2)(cos(t)+i*sin(t))^(3i)

=5(cos(2t)+i*sin(2t)*(5^(3/2)(cos(3t) + i*sin(3t))^i

The only tricky thing left is how to calculate z^i where z is complex.

Let z^i = w. Then i*ln(z) = ln(w). ln(z) can be calculated using the exponential form. ln(Re^(it)) = ln(R) + it since exp and ln are inverse functions. and ln(w) = -t+i*ln(R), so w = (e^-t)*e^(i*ln(R))

= e^(-t)*(cos(ln(R) + i*sin(ln(R)).

So (5^(3/2)(cos(3t) + i*sin(3t))^i becomes e^(-3t)*(cos(ln(5^(3/2)))+i*sin(ln(5^(3/2))))

and our final expression 5(cos(2t)+i*sin(2t)*e^(-3t)*(cos(ln(5^(3/2)))+i*sin(ln(5^(3/2))))

This gives the same as my TI92.

Best wishes

vijilant

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Comment | Dear Prof Vijilant Thanks Prashant |

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