what is the value of log sqrt 3i?

Answer 1

The other answer is off just a little. He should have taken the natural log of each factor of √(3i) and added them.

√(3i) = √3 * √i

So ln [√(3i)] = ln [√3 * √i] = ln (√3) + ln (√i) = ½ ln (3) + ½ ln (i).

Now you can back substitute the value for i:

ln [√(3i)] =
½ ln (3) + ½ ln (i) =
½ ln (3) + ½ ln [e^(iπ/2)] =
½ ln (3) + ½ (iπ/2) =
½ [ln (3) + (iπ/2)]

Answer 2

Remember that i can be expressed as e^(i pi/2).

So 3i is 3e^( i pi/2). The square root of that is

sqrt (3i) = (3i)^1/2 = sqrt(3)e^(i pi/4)

And taking the log (I assume you mean natural log) is thus simply the exponent of e, i.e.

log sqrt 3i = log (sqrt(3)) + i pi/4 = (log 3)/2 + i pi/4