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Expert: - 9/23/2010


Question
Hi, may i get some leads on how to go about this question please? googled lecture notes and txtbooks but couldnt find an exponential to the power of a complex number.

Qn: By taking logarithm, solve the eqn exp(2+3iz) = 2i
please correct me if i have made a mistake in my initial steps.

i took natural log both sides, getting me 2+3iz = ln(2i), making z the subject, i get z = (-i/3)(ln 2i - 2), i'm kind of stuck here now.

and also, express sin(3+i) in cartesian form.

theres alot more qns that branch from here, once i get these techniques right, i hope i can do the rest myself. thank you so much!

Answer
I referred to www.kith.org/logos/things/euler.html

I also used the fact fact the ln(ab) = ln(a) + ln(b)
and 2ln(a)/2 = ln(aČ)/2.

Using these, it can be seen that if we have 2 + 3iz = ln(2i),
that is really 2 + 3iz = ln(2) + ln(i).

Now ln(i) = ln(iČ)/2 = ln(-1)/2, so we get the equation
2 + 3iz = ln(2) + ln(-1)/2.

Solving for z means that first we subtract 2 from both sides, giving
3iz = ln(2) + ln(-1)/2 - 2.

We then multiply both sides by -i/3, giving
z = -i(ln(2) + ln(-1)/2 - 2)/3.

As in the paper, ln(-1) = i, so we have z = -i(ln(2) + i/2 - 2)/3.

Finishing this process means the middle term gives the real part and the first and last give the imaginary part.  That is,
z = (-1/2 - i(ln(2) - 2))/3 = -1/6 - i(ln(2) - 2)/3 .


For trig, see math.stanford.edu/~clingher/spring06/116hw4.pdf

To apply the sum of two numbers in trig, use the trig identity.
For sin(a+b) we get sin(a)cos(b) + sin(b)cos(a).
For his problem, use a=3 and b=i.

In general,
sin(i) = (e^i - e^(-i))/2 and
cos(i) = (e^i + e^(-i))/2.

This leads to
tan(1+i) = (cos1*sini + sin1*cosi)/(cos1*cosi − sin1*sini).

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