Correct to three decimal places, find the real part of the complex number: cos ( 1 + i ) .
You may need a scientific calculator to solve this problem.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
nice solution ,upvoted
Yup, same solution.
I think you forgot to put the i at the sin ( x ) sinh ( y ) part
Log in to reply
You're right; thanks for pointing that out. :)
By Angle-Sum Identity: cos ( 1 + i ) = cos 1 cos i − sin 1 sin i
cos x = 2 e i x + e − i x ⇒ cos i = 2 e i 2 + e − i 2 = 2 e − 1 + e = cosh 1 sin x = 2 i e i x − e − i x ⇒ sin i = 2 i e i 2 − e − i 2 = 2 i e − 1 − e = − 2 i e − e − 1 = i sinh 1
∴ cos ( 1 + i ) = cos 1 cosh 1 − i sin 1 sinh 1 ⇒ a = R e a l P a r t = cos 1 cosh 1 = 0 . 8 3 4
Hint → Angles in radians .
Forgot to put angle in radians and got all three tries wrong :p
Why the 1 in cos 1 have to be in radians?
Log in to reply
Because it wasn't written as being in degrees. If there aren't any symbols around an angle measure (like 1), it's usually in radians.
We can write c o s ( x ) = 2 e i x + e − i x , derived from Euler's formula
c o s ( 1 + i ) = 2 e − 1 + i + e 1 − i
Let y = e − 1 + i = e − 1 ∗ e i
Now, Euler's formula states that e i x = c o s ( x ) + i s i n ( x ) ; thus, e i = c o s ( 1 ) + i ∗ s i n ( 1 ) ; therefore, y = e c o s ( 1 ) + i ∗ s i n ( 1 ) , and from this we conclude that c o s ( 1 + i ) = 2 y + y 1 = 0 . 8 3 3 7 3 0 0 2 5 − 0 . 9 8 8 8 9 7 7 0 6 i .
The problem asks for the real part of c o s ( 1 + i ) , which equals approximately 0 . 8 3 3 .
Try Maclaurin series of cos(x)=1-(x^2)/2!+(x^4)/4!+.... Also, (1+i)^2=2i, so (1+i)^4=-4. So all terms that contain powers of x^4 are real. So the desired answer is 1-4/4!+16/8!-64/12!+... Since the problem wants 3 decimal places, and the first factorial in the desired series to exceed 1000 is 8!, the answer is 1-4/4!+16/8! truncated, or 0.833.
Nice solution!
I used the series
cos ( z ) = n ∑ ( − 1 ) n ⋅ z 2 n / ( 2 n ) !
Letting z = 1 + i = 2 ⋅ e i ⋅ π / 4 , we have
cos ( z ) = n ∑ ( − 1 ) n ⋅ 2 n ⋅ e i ⋅ n π / 2 / ( 2 n ) !
Now,
e i ⋅ n π / 2 = { ( − 1 ) m ( − 1 ) m ⋅ i i f $ n = 2 m $ i f $ n = 2 m + 1 $
and therefore:
R e cos ( z ) = m ∑ ( − 1 ) m ⋅ 4 m / ( 4 m ) ! = 1 − 4 ! 4 + 8 ! 1 6 + ⋯ ≈ 0 . 8 3 4
The Alternating Series Error Rule tells us that two terms will suffice (if both truncating and rounding are allowed), since 8 ! 1 6 < 0 . 0 0 0 5 .
Can we use Taylor Series here???I did the same and got the answer.
Definitely! TS apply equally in R and C. 😉
Just google,
cos(1+i)
;) - simple
Than what's the point of being in Brilliant Community.. You guys should be removed..This is a maths community.. All the problems are meant to be solved..Not copying from google..
Log in to reply
what about graphing r=cos(i+1), am i booted out of the club
Thts unethical on brilliant buddy 😁
Problem Loading...
Note Loading...
Set Loading...
By definition, for z ∈ C , we have that cos ( z ) = 2 1 ( e i z + e − i z ) .
So in general, for x , y ∈ R , we have that
cos ( x + i y ) = 2 1 ( e i ( x + i y ) + e − i ( x + i y ) ) = 2 1 ( e − y + i x + e y − i x ) =
2 1 [ e − y ( cos ( x ) + i sin ( x ) ) + e y ( cos ( x ) − i sin ( x ) ) ] =
2 1 ( cos ( x ) ( e y + e − y ) − i sin ( x ) ( e y − e − y ) ) = cos ( x ) cosh ( y ) − i ∗ sin ( x ) sinh ( y ) ,
where the definitions cosh ( y ) = 2 e x + e − x and sinh ( y ) = 2 e y − e − y were used.
In this case we have x = y = 1 , and so a = cos ( 1 ) cosh ( 1 ) = 0 . 8 3 3 7 3 to 5 decimal places.