tan(x + iy), and Separating It Into Real and Imaginary Parts
Chapter Thirty-Three
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 74 to 75 of 303
In one line
Separate the tangent of x + iy by multiplying above and below by the conjugate of the denominator, exactly as you divide any complex number.
Why it is harder than the sine and the cosine
The tangent is a quotient, so separating it means dividing one complex number by another, and the answer's parts are both fractions. It is the single most demanding routine manipulation in Module 1, and it is set regularly, so it gets a chapter.
There are two routes. The first is shorter and the second is safer, and it is worth being able to do both.
Route one: from the sine and the cosine
Write the tangent as the quotient, substitute the two separations from the earlier chapter, and then divide by the standard method.
tan(x + i y) = sin(x + i y)/cos(x + i y)
sin(x + i y) = sin(x) cosh(y) + i cos(x) sinh(y)
cos(x + i y) = cos(x) cosh(y) - i sin(x) sinh(y)
Now multiply above and below by the conjugate of the bottom, which is cos x cosh y plus i sin x sinh y. The new denominator is the modulus squared of the cosine, which the earlier chapter showed to be cos squared x plus sinh squared y.
abs(cos(x + i y))^2 = cos(x)^2 + sinh(y)^2
The numerator, multiplied out and with cosh squared minus sinh squared equals one used on it, gives the following.
tan(x + i y) = (sin(x) cos(x) + i sinh(y) cosh(y))/(cos(x)^2 + sinh(y)^2)
So the real part is sin x cos x over that denominator and the imaginary part is sinh y cosh y over the same denominator.
The tidier form, which is the one to quote
Use the double-angle formulae on both numerator terms and on the denominator. Twice sin x cos x is sin 2x, twice sinh y cosh y is sinh 2y, and twice the denominator is cos 2x plus cosh 2y.
2 sin(x) cos(x) = sin(2x)
2 sinh(y) cosh(y) = sinh(2y)
2(cos(x)^2 + sinh(y)^2) = cos(2x) + cosh(2y)
Doubling top and bottom therefore gives the standard result.
tan(x + i y) = (sin(2x) + i sinh(2y))/(cos(2x) + cosh(2y))
That is the form worth carrying. It is symmetric, it has one denominator, and both parts are immediate.
Route two: from the compound-angle formula for the tangent
Use the tangent addition formula directly, with the relation that the tangent of an imaginary angle is i times a hyperbolic tangent.
tan(i y) = i tanh(y)
tan(x + i y) = (tan(x) + i tanh(y))/(1 - i tan(x) tanh(y))
Then multiply above and below by the conjugate of the denominator and collect. The answer is the same, and this route is quicker if the question is already in terms of tangents. Its drawback is that it breaks down where tan x is undefined, at odd multiples of pi over two, which the first route handles without difficulty.
tan(x + iy), and Separating It Into Real and Imaginary Parts
The hyperbolic companion
The same work on tanh, either by Osborn's rule applied to the result or from the definitions, gives the following.
tanh(x + i y) = (sinh(2x) + i sin(2y))/(cosh(2x) + cos(2y))
The roles of x and y and of the two families have swapped, exactly as they did for the sine and cosine pair.
Worked, with numbers
Separate tan(pi over 4 plus i) into real and imaginary parts.
Here x is pi over four and y is one. So 2x is pi over two, and sin 2x is one while cos 2x is zero.
tan(pi/4 + i) = (sin(pi/2) + i sinh(2))/(cos(pi/2) + cosh(2))
tan(pi/4 + i) = (1 + i sinh(2))/cosh(2)
So the real part is sech 2, about 0.2658, and the imaginary part is tanh 2, about 0.9640.
1/cosh(2) = sech(2)
sinh(2)/cosh(2) = tanh(2)
A second one, where the angle is purely imaginary. Put x = 0 and the formula collapses.
tan(i y) = (0 + i sinh(2y))/(1 + cosh(2y))
i sinh(2y)/(1 + cosh(2y)) = i tanh(y)
which is the relation the previous chapter gave, arrived at from the general formula. That agreement is a good check that the general formula has been remembered correctly.
The procedure
- Identify x and y.
- Write the standard result, with sin 2x and sinh 2y on top and cos 2x plus cosh 2y underneath.
- Evaluate the four trigonometric and hyperbolic values.
- Read off the two parts.
- Check the special cases: y = 0 must give tan x, and x = 0 must give i tanh y.
Step five is the one that catches a misremembered formula in ten seconds.
(sin(2x) + i sinh(0))/(cos(2x) + cosh(0)) = tan(x)
Putting y = 0 does return the ordinary tangent, so the formula is right.
Check yourself
tan(x + i y) = (sin(2x) + i sinh(2y))/(cos(2x) + cosh(2y))
tanh(x + i y) = (sinh(2x) + i sin(2y))/(cosh(2x) + cos(2y))
tan(i) = i tanh(1)
tan(pi/4) = 1
abs(tan(i y)) = abs(tanh(y))
The fourth line is the y = 0 check on the standard formula, and the last says the tangent of an imaginary angle has modulus at most one, because tanh does. That is the opposite of what happens to the sine and cosine, which grow without limit up the imaginary axis.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.