Find the value of (a + bi)^(c + di)?

Question

Find the value of (a + bi) raised to the power (c + di)

^_^

Answer 1

You can rewrite this as (a + bi)^c * (a + bi)^di. It's much easier to calculate if we rewrite a + bi in exponential form. That's R*e^T, where R is the radius or modulus, sqrt(a^2 + b^2), and T (for theta) is the angle or argument, arctan(b/a). Then recall that e^iT = cos(T) - i*sin(T). To get rid of an R^i term, rewrite R as e^ln(R), and use the trig expansion. The end result is R^c * e^-dT * [cos(d*ln(R) + cT) + i*sin(d*ln(R) + cT)], with R and T defined above.

Answer 2

The way this question is worded is confusing. By the way I interpret it, the solution would seem to be itself. Like asking what is the value of 3? The value of 3 is 3. So therefore the value of (a + bi)^(c + di) would equal (a + bi)^(c + di), unless you're given any additional information and asked to solve for a specific variable.

The only other thing I can think of doing is changing the i's to sqrt(-1)'s, as sqrt(-1) = i. Beyond that, don't know what to tell you.

Answer 3

No solution

(a + bi)ˆ(c + di) requires numerical values

Answer 4

O_O
I don't think there's a general form for it, so unless you give values for a, b, c, and d, I doubt it's possible.

Answer 5

(a+bi)^(c+di) = e^{ (c+di) ln(a+bi) }
= e^{ (c+di) ln[ r(cosP+i sinP)] }
= e^{ (c+di) ln[ re^(iP) ] }
= e^{ (c ln r - dP) + i( d ln r + cP ) }

There's a lot more info at the page I've linked to below. Hope that helps!

Answer 6

(a + bi)^(c + di)
((a + bi)^(c)) * ((a + bi)^(di))

Thats all i can tell you

Answer 7

KEVIN, R U JOKING?
I MEAN REALLY!!! HOW WOULD EXPECT US TO FIND A VALUE? IT'S NOT EVEN AN EQUATION!!!!

CAN ANYBODY SHOW ME HOW , IF ITS POSSIBLE,....THAT I GOTTA SEE!!!!