Question

Convert all complex numbers to trigonometric form and then simplify the expression? Write the answer in standard form.

`((2 + 2i)^5(-3+i)^3)/((sqrt3 +i)^10`

Answer 1

`{ (2+2i)^5 (-sqrt{3} + i)^3 }/(sqrt{3}+i)^10`

`= (sqrt{3}-1)/2 + (sqrt{3}+1 )/2 \ i`

Explanation:

As anyone who reads my answers may have noticed, my pet peeve is every trig problem involves a 30/60/90 or 45/45/90 triangle. This one has both, but `-3+i` is neither.

I'm going to go out on a limb and guess the question in the book actually read:

Use trigonometric form to simplify

`{ (2+2i)^5 (-sqrt{3} + i)^3 }/(sqrt{3}+i)^10`

because this way would only involve the Two Tired Triangles of Trig.

Let's convert to trigonometric form, which is just polar form written

`r\ text{cis}\ theta = r(\cos theta + i \sin theta)`

Then by De Moivre's Thorem

`(r\ text{cis}\ theta )^n = r^n \ text{cis}(n theta)`

Let's convert each factor.

`|2+2i|=sqrt{2^2+2^2}=2sqrt{2}`

`angle 2+2i = 45^circ`

`2 + 2i = 2 \sqrt{2} \ text{cis}\ 45^circ`

`|-sqrt{3} + i| = sqrt{ (-sqrt{3})^2 + 1^2}=2`

`angle (-sqrt{3} + i) = 150^circ`

That's the 30/60/90 triangle in the second quadrant whose cosine is bigger (in magnitude) than its sine.

`-sqrt{3} + i = 2 \ text{cis}\ 150^circ`

Similarly,

`\sqrt{3} + i = 2 \ text{cis}\ 30^circ`

Now,

`{ (2+2i)^5 (-sqrt{3} + i)^3 }/(sqrt{3}+i)^10`

`= { ( 2 \sqrt{2} \ text{cis}\ 45^circ )^5 ( 2 \ text{cis}\ 150^circ )^ 3 }/{ ( 2 \ text{cis}\ 30^circ ) ^10 }`

`= { (2 \sqrt{2})^5 \ text{cis}\ (5 cdot 45^circ )( 2^3 \ text{cis}\ ( 3 cdot 150^circ) ) }/ { 2^10 \ text{cis}\(10 cdot 30^circ ) }`

#= {2^5 \sqrt{2^5} \ 2^3 }/ 2^{10} {\ text{cis}\ 225^circ \ text{cis}\ 450^circ }/{\ text{cis}\ 300^\circ } #

`= sqrt{2} \ text{cis}\ (225^circ + 450^circ - 300 ^circ )`

`= \sqrt{2} \ text{cis}\ 375^circ`

`= \sqrt{2} \ text{cis}\ 15^circ`

I don't want to deal with the trig functions of `15^circ,` which are best gotten from the difference angle formulas with `15^circ = 45^circ - 30^circ,` our tired triangles. We can avoid it by redoing the problem, cleverly leaving one factor of `2+2i` till the end:

`{ (2+2i)^5 (-sqrt{3} + i)^3 }/(sqrt{3}+i)^10`

`= (2+2i) { (2 \sqrt{2})^4 \ text{cis}\ (4 cdot 45^circ )( 2^3 \ text{cis}\ ( 3 cdot 150^circ) ) }/ { 2^10 \ text{cis}\(10 cdot 30^circ ) }`

`= (2+2i) {2^4 \sqrt{2^4} \ 2^3}/ 2^{10} {\ text{cis}\ 180^circ \ text{cis}\ 450^circ }/{\ text{cis}\ 300^\circ }`

#= (2+2i)/2 \ text{cis}\ 30^circ #

`= (1+i)cdot (\sqrt{3} + i)/2`

`= (sqrt{3}-1)/2 + (sqrt{3}+1 )/2 \ i`

Answer 2

`{11 + 2 sqrt(3)}/4 + (11 sqrt(3) - 2)/4 i`

Explanation:

In another answer to this question I guessed there was a typo in this question and that `-3` was supposed to be `-sqrt{3}`. I've been assured in a comment that that's not the case, that the question is correct as written.

I won't repeat how we determined

`2+ 2i = 2 sqrt{2} \ \text{cis}\ 45^circ`

`sqrt{3} + i = 2 \ \text{cis}\ 30^circ`

But now we have to convert `-3 + i` to trigonometric form. We can do it, but since it's not one of Trig's preferred triangles, it's a bit more awkward.

`|-3 + i|=sqrt{3^2+1^2}=sqrt{10}`

We're in the second quadrant and the principal value of the inverse tangent is the fourth quadrant.

`\angle (-3 + i) = \text{Arc}text{tan} (1 /{-3}) + 180^circ`

`-3 + i = \sqrt{10} \ \text{cis} ( text{Arc}text{tan} (1 /{-3}) + 180^circ )`

De Moivre doesn't work very well on a form like this, we get

`(-3+i)^3 = \sqrt{10^3 } \ \text{cis} ( 3 ( text{Arc}text{tan} (1 /{-3}) + 180^circ ) )`

But we're not stuck. Since the exponent is only `3` we can do this with triple angle formulas. Let's call the constant angle we found

`theta = angle (-3 + i)`

By De Moivre,

`(-3+i)^3 = (\sqrt{10 } \ \text{cis} theta )^3 = 10sqrt{10} (\cos (3theta) + i \sin (3 theta) )`

We know

`\cos theta = -3/sqrt{10}, quad sin theta = 1/sqrt{10}`

`cos(3 theta) = 4 cos ^3 theta - 3 cos theta = 4(-3/sqrt{10})^3 - 3(- 3 /sqrt{10}) = -(9 sqrt(10))/50`

`sin(3 theta) = 3 sin theta - 4 sin ^3 theta = 3( 1/sqrt{10} ) - 4( 1/sqrt{10} )^3 = (13 sqrt(10))/50`

`(-3+i)^3 =10sqrt{10} (sqrt{10}/50) (-9 + 13 i ) = -18 +26 i`

That seems like much more work than just cubing `(-3+i):`

`(-3+i)(-3+i)(-3+i)=(-3 + i)(8 -6i) = -18 + 26 i quad sqrt`

OK, let's do the problem:

`{( 2 + 2i)^5 (-3 + i)^3 }/{( sqrt{3} + i )^{10} }`

`= {( 2 sqrt{2} \ \text{cis}\ 45^circ )^5 (-3 + i)^3 }/{( 2 \ \text{cis}\ 30^circ)^{10} }`

# = ( { 2^5 sqrt{2^5 }}/2^10 ) { \text{cis}(5 cdot 45^circ ) }/{\text{cis}(10 cdot 30^circ) } (-3 + i)^3 #

# = (sqrt{2}/8) { \text{cis}(225^circ ) }/{\text{cis}( 300^circ) } (-3 + i)^3 #

`= (sqrt{2}/8) \text{cis}(225^circ - 300) (-3 + i)^3`

`= (sqrt{2}/8) ( -18 +26 i ) \text{cis}(-75^circ)`

Ugh, it never ends. We get

`cos(-75^circ)=cos 75^circ = cos(45^circ + 30^circ) = \sqrt{2}/2(sqrt{3}/2 - 1/2) = 1/4(sqrt{6} -sqrt{2})`

`sin(-75^circ) = -( sin 45 cos 30 + cos 45 sin 30) = -sqrt{2}/2 (\sqrt{3}/2 + 1/2)=-1/4(\sqrt{6} + sqrt{2})`

`{( 2 + 2i)^5 (-3 + i)^3 }/{( sqrt{3} + i )^{10} }`

`= (sqrt{2}/8) ( -18 +26 i ) 1/4 ( (sqrt{6} -sqrt{2} ) - (sqrt{6}+sqrt{2})i)`

`= {11 + 2 sqrt(3)}/4 + (11 sqrt(3) - 2)/4 i`