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- WELCOME TO A LESSON ON DE MOIVRE'S THEOREM.
THE GOAL IS TO USE DE MOIVRE'S THEOREM
TO RAISE COMPLEX NUMBERS TO ANY POWER.
SO DE MOIVRE'S THEOREM
MAKES RAISING COMPLEX NUMBERS TO POWERS VERY SIMPLE.
IF R x THE QUANTITY COSINE THETA + "I" SINE THETA
IS A COMPLEX NUMBER, AND IF N IS ANY REAL NUMBER,
THEN WE TAKE THIS COMPLEX NUMBER AND RAISE IT TO THE POWER OF N,
IT'S EQUAL TO R TO THE POWER OF N
x THE QUANTITY COSINE N THETA + "I" SINE N THETA.
SO THIS IS A VERY INTERESTING RESULT.
THIS VALUE OF N IS USED FOR THE POWER OF R,
BUT THEN IT'S MULTIPLIED BY THETA TO FIND THE NEW ANGLE.
LET'S SEE IF WE CAN DETERMINE WHERE THIS FORMULA COMES FROM.
ONE WAY WOULD BE TO JUST START
WITH A COMPLEX NUMBER IN TRIG FORM AND SQUARE IT,
OR MULTIPLY IT BY ITSELF.
THIS WILL REMIND YOU OF THE PRODUCT
OF COMPLEX NUMBERS IN TRIG FORM.
OF COURSE, WE'D HAVE R x R OR R SQUARED.
THEN WHEN WE MULTIPLY THIS OUT IT WOULD BE THE SAME THING
AS WE SAW IN THE PREVIOUS VIDEO FOR THE PRODUCT,
EXCEPT NOW THE TWO ANGLES ARE THE SAME.
SO WHEN WE APPLY THIS SUM IDENTITY FOR A COSINE,
SINCE ANGLE "A" AND ANGLE B WOULD BE THE SAME,
WE'D HAVE THETA + THETA OR 2 THETA.
SO WE'D HAVE THE COSINE OF 2 THETA HERE,
AND THE SAME FOR THE SUM IDENTITY FOR SINE,
SINCE THE ANGLES "A" AND B WOULD BE THE SAME.
WE WOULD HAVE THE SINE OF THETA + THETA OR THE SINE OF 2 THETA.
IF WE TAKE THIS Z SQUARED AND MULTIPLY IT BY ANOTHER Z,
AS WE SEE HERE, WE'LL END UP WITH R CUBED,
AND THEN THE COSINE OF 3 THETA + "I" SINE 3 THETA.
SO WHAT WE'RE NOTICING HERE
IS THAT THE POWER THAT Z IS RAISED TO
IS USED ON THE POWER OF R,
BUT THEN IT'S MULTIPLIED BY THETA TO FIND THE NEW ANGLE.
AND THIS PATTERN DOES CONTINUE.
ANOTHER WAY TO TAKE A LOOK AT THIS THEOREM
WOULD BE TO USE EULER'S THEOREM,
WHICH STATES E TO THE POWER OF "I" THETA
= COSINE THETA + "I" SINE THETA.
SO OF COURSE IT WOULD FOLLOW THAT R, E TO THE "I" THETA,
= R x COSINE THETA + "I" SINE THETA.
BEFORE WE SHOW THE NEXT STEP,
WHAT WE SHOULD RECOGNIZE HERE IS THAT OUR ARGUMENT, THETA,
ENDS UP AS THE ANGLE FOR SINE AND COSINE.
SO NOW IF WE TAKE THIS AND RAISE IT TO THE POWER OF N,
THIS IS R TO THE FIRST POWER.
SO WE WOULD HAVE R TO THE N, E TO THE "I" N THETA.
WELL, NOW OUR ARGUMENT IS N THETA INSTEAD OF THETA,
SO NOW WE COULD SAY THIS WOULD BE EQUAL TO R TO THE N
x COSINE N THETA + "I" SINE N THETA.
AND THIS IS DE MOIVRE'S THEOREM,
THAT STATES THAT THIS RAISED TO THE POWER OF N
= R TO THE N x THE QUANTITY COSINE N THETA
+ "I" SINE N THETA.
LET'S GO AHEAD AND SEE IF WE CAN APPLY THIS.
WE WANT TO REWRITE THIS IN RECTANGULAR FORM,
BUT WHAT WE'LL DO FIRST IS EVALUATE THIS
USING DE MOIVRE'S THEOREM,
AND THEN WE'LL CONVERT THIS TO RECTANGULAR FORM.
SO THIS WOULD BE THE SAME AS 2 TO THE 3rd POWER
x THE COSINE OF 3 x 3PI/4 + "I" SINE OF 3 x 3PI/4.
SO WE CAN SEE OUR ANGLE IS GOING TO BE 9PI/4,
SO WE'LL HAVE 8 x COSINE 9PI/4 + "I" SINE 9PI/4.
LET'S TAKE A LOOK AT THE ANGLE 9PI/4.
9PI/4 WOULD BE COTERMINAL WITH AN ANGLE OF 9PI/4 - 2PI.
WELL, 2PI IS THE SAME AS 8PI/4, SO THIS WOULD EQUAL PI/4.
SO THIS ANGLE IS COTERMINAL WITH PI/4,
SO WE'LL USE THAT AS OUR REFERENCE ANGLE.
THIS IS 45 DEGREES IN THE FIRST QUADRANT.
SO WE HAVE 8 x THE COSINE OF 9PI/4.
THAT WOULD BE 1/SQUARE ROOT 2,
OR SQUARE ROOT 2/2 IF WE RATIONALIZE IT,
AND THE SAME FOR THE SINE OF 9PI/4, 1/SQUARE ROOT 2,
OR RATIONALIZED WOULD BE SQUARE ROOT 2/2.
SO WE HAVE 4 SQUARE ROOT 2 + 4I SQUARE ROOT 2.
SO YOU CAN SEE IF WE CONVERTED IT TO RECTANGULAR FORM FIRST
AND THEN TRIED TO CUBE IT, IT WOULD BE A LOT MORE WORK.
NOW WE HAVE A COMPLEX NUMBER
IN RECTANGULAR FORM RAISED TO THE 5th POWER,
AND WE WANT TO USE DE MOIVRE'S THEOREM TO COMPUTE THIS.
SO WE'LL CONVERT THIS TO TRIG FORM,
AND THEN WE'LL RAISE IT TO THE 5th POWER.
NOW I INCLUDE THE UNIT CIRCLE HERE,
BECAUSE IT'S GOING TO HELP US DRAMATICALLY.
REMEMBER, FOR THIS COMPLEX NUMBER "A" = 1/2
AND B = -SQUARE ROOT 3/2.
EVEN THOUGH THIS IS THE RECTANGULAR COORDINATE PLANE
RATHER THAN THE POLAR COORDINATE PLANE,
WE CAN USE THIS TO DETERMINE THETA.
REMEMBER, ON THE RECTANGULAR COORDINATE PLANE
"A" WOULD BE X AND B WOULD BE Y.
SO WHEN X IS 1/2 AND Y IS -SQUARE ROOT 3/2,
WE'RE ACTUALLY RIGHT HERE,
WHERE OUR ANGLE IS 300 DEGREES OR 5PI/3 RADIANS.
SO THAT'S HELPFUL. LET'S USE 300 DEGREES.
NEXT WE'RE FORTUNATE,
BECAUSE THAT'S ACTUALLY THE POINT ON THE UNIT CIRCLE,
THEREFORE R = 1.
BUT REMEMBER THIS IS ALL WE NEED
TO WRITE THIS IN TRIGONOMETRIC FORM.
WE HAVE 1 x THE COSINE OF 300 DEGREES + "I" SINE 300 DEGREES,
BUT WE'RE TAKING THIS AND RAISING IT TO THE 5th POWER.
SO USING DE MOIVRE'S THEOREM
WE'LL HAVE 1 TO THE 5th x THE COSINE OF 5 x 300
= 1,500 DEGREES + "I" SINE 1,500 DEGREES.
LET'S GO AHEAD AND TAKE A LOOK AT 1,500 DEGREES.
LET'S FIND THE LEAST POSITIVE ANGLE
THAT IS COTERMINAL TO THAT.
SO WHAT WE'LL DO IS WE'LL TAKE 1,500 DEGREES, DIVIDE BY 360,
SO WE HAVE 4, AND THAT'S 1,440, WE HAVE A REMAINDER OF 60,
SO THAT TELLS US THAT THIS IS COTERMINAL
TO A POSITIVE 60-DEGREE ANGLE.
SO SINCE WE ALREADY HAVE THE UNIT CIRCLE HERE,
WE'RE ACTUALLY GOING TO USE THIS POINT RIGHT HERE
TO DETERMINE THESE FUNCTION VALUES.
WELL, 1 TO THE 5th = 1, COSINE OF 1,500 DEGREES,
REMEMBER ON THE UNIT CIRCLE THAT WOULD BE THE X COORDINATE,
SO WE HAVE 1/2 + "I" x THE SINE OF 1,500 DEGREES.
WELL, SINE WOULD BE THE Y COORDINATE,
SO WE HAVE SQUARE ROOT 3/2 x "I", OR "I" SQUARE ROOT 3/2.
SO THERE WE HAVE IT.
THIS IS THAT COMPLEX NUMBER RAISED TO THE 5th POWER.
SO AS YOU CAN SEE, AGAIN,
IF WE TRIED TO RAISE THIS TO THE 5th POWER IN RECTANGULAR FORM
THAT WOULD BE VERY DIFFICULT TO DO,
BUT ONCE WE CONVERT IT TO TRIG FORM
AND USE DE MOIVRE'S THEOREM,
IT ACTUALLY IS VERY STRAIGHT- FORWARD AND A LOT EASIER.
I HOPE YOU FOUND THIS VIDEO HELPFUL. HAVE A GOOD DAY.