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? asked in Science & MathematicsMathematics · 1 decade ago

Real number raised to complex number?

so Euler's formula make e^ix=cosx+isinx..where Euler get this answer?

my question is is it make sense let a real number raised to complex number

for example like: 7^(10+i8),and how should i compute this one?

or let a complex number raised to a real number,

for example (10+i7)^8,and how should i compute this one?

or even let a complex number raised to complex number;

for example: (10+i7)^(13-i8), and how should i compute this one?

2 Answers

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  • 1 decade ago
    Favorite Answer

    7^(10 + 8i) = 7^10 * 7^(8i)

    7^10 = 282,475,249

    7^(8i) = [e^(ln 7)]^(8i)

    = e^[8 (ln 7) i]

    = cos(8 ln 7) + i sin(8 ln 7)

    = approx. cos(15.57) + i sin(15.57)

    = approx. 0.9633 + 0.2684 i

    So 7^(10 + 8i) = approx. 282,475,249 * (0.9633 + 0.2684 i)

    = approx. 272,112,920 + 75,807,815 i

    (10 + 7i)^2 = 100 - 49 + 140i = 51 + 140i

    (10 + 7i)^4 = (51 + 140i)^2 = - 16,999 + 14,280i

    (10 + 7i)^8 = (- 16,999 + 14,280i)^2

    = 85,047,601 - 485,491,440i

    (10 + 7i)^(13 - 8i)

    = (10 + 7i)^13 / (10 + 7i)^(8i)

    (10 + 7i)^13 can be evaluated as in question 2.

    10^2 + 7^2 = 100 + 49 = 149

    arctan(7/10) = approx. 0.6107

    so 10 + 7i = approx. (sqrt 149) e^(0.6107i)

    and (10 + 7i)^(8i) = approx. [(sqrt 149) e^(0.6107i)]^(8i)

    = approx. (sqrt 149)^(8i) * e^(- 0.6107 * 8)

    (sqrt 149)^(8i) can be evaluated by a procedure similar to that of the question 1.

    e^(- 0.6107 * 8) is a real number.

    So (10 + 7i)^(13 - 8i) can be evaluated, but I think I'll leave it to you to find its value!

  • 5 years ago

    you could coach that i^i is real very in the present day utilising merely logs. when you consider that i = e^(-i ?/2), ln(i) = i ?/2. Substituting: ln(i^i) = i ln(i) = i (i ?/2) = -?/2 So i^i = e^(-?/2) There are quite a few coaching on a thank you to coach that i^i is real yet you already be attentive to that. What i think of is greater exciting is to visualise what is going on. First, i is imaginary with a magnitude of one million. to accomplish an operation on it to transform it to a real demands a rotation interior the complicated polar plane of -?/2. For i^i to be real, all you go with is that the exponentiation does rotate i via -?/2. you be attentive to that a rotation of -?/2 is comparable to multiplying via e^(-i?/2). So i*e^(-i?/2) is real and the magnitude of one million is preserved. The expression i^i = i*e^(-i?/2) has the variety i*e^c and we choose an equivalent f*i^d to discover the exponent that does the rotation. so a ways all of us be attentive to c yet ought to unravel for f and d. i've got further an element of f because of the fact so a ways i've got dealt with rotation yet disregarded magnitude. i choose one greater unknown to account for the reality that the magnitude of i^i choose not be one million. merely take the log of the two varieties and set them equivalent: Taking the log: ln(i*e^c) = ln(i) + c = ln(f) + d*ln(i) fixing for ln(f): ln(f) = ln(i) (one million - d) + c all of us be attentive to c = -i ?/2 of direction i = e^(i ?/2) so ln(i) = i ?/2 so: ln(f) = (i?/2)(one million - d) - (i?/2) = id?/2 when you consider that f is basically a magnitude, it shall be real and the log must be real so id?/2 must be real so d is imaginary. So any imaginary exponent of i will produce a real result. back, you should coach this result greater honestly via merely dong the logs.

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