Elementary Complex Number

Algebra Level 5

Correct to three decimal places, find the real part of the complex number: cos ( 1 + i ) \cos (1 + i ) .

You may need a scientific calculator to solve this problem.


The answer is 0.833.

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7 solutions

By definition, for z C z \in \mathbb{C} , we have that cos ( z ) = 1 2 ( e i z + e i z ) . \cos(z) = \dfrac{1}{2}(e^{iz} + e^{-iz}).

So in general, for x , y R x,y \in \mathbb{R} , we have that

cos ( x + i y ) = 1 2 ( e i ( x + i y ) + e i ( x + i y ) ) = 1 2 ( e y + i x + e y i x ) = \cos(x + iy) = \dfrac{1}{2}(e^{i(x + iy)} + e^{-i(x + iy)}) = \dfrac{1}{2}(e^{-y + ix} + e^{y - ix}) =

1 2 [ e y ( cos ( x ) + i sin ( x ) ) + e y ( cos ( x ) i sin ( x ) ) ] = \dfrac{1}{2}[e^{-y}(\cos(x) + i\sin(x)) + e^{y}(\cos(x) - i\sin(x))] =

1 2 ( cos ( x ) ( e y + e y ) i sin ( x ) ( e y e y ) ) = cos ( x ) cosh ( y ) i sin ( x ) sinh ( y ) \dfrac{1}{2}(\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 ) = e x + e x 2 \cosh(y) = \dfrac{e^{x} + e^{-x}}{2} and sinh ( y ) = e y e y 2 \sinh(y) = \dfrac{e^{y} - e^{-y}}{2} were used.

In this case we have x = y = 1 x = y = 1 , and so a = cos ( 1 ) cosh ( 1 ) = 0.83373 a = \cos(1)\cosh(1) = \boxed{0.83373} to 5 decimal places.

nice solution ,upvoted

Utkarsh Bansal - 6 years, 3 months ago

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Nice question @Utkarsh Bansal ...

Parth Lohomi - 6 years, 3 months ago

Yup, same solution.

Jake Lai - 6 years, 3 months ago

I think you forgot to put the i i at the sin ( x ) sinh ( y ) \sin{(x)}\sinh{(y)} part

Julian Poon - 6 years, 3 months ago

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You're right; thanks for pointing that out. :)

Brian Charlesworth - 6 years, 3 months ago
Mustafa Embaby
Feb 17, 2015

By Angle-Sum Identity: cos ( 1 + i ) = cos 1 cos i sin 1 sin i \\ \cos(1+i)=\cos 1\cos i-\sin 1\sin i


cos x = e i x + e i x 2 cos i = e i 2 + e i 2 2 = e 1 + e 2 = cosh 1 \cos x = \dfrac {e^{ix} + e^{-ix}}{2} \Rightarrow \cos i = \dfrac {e^{i^2} + e^{-i^2}}{2} = \dfrac {e^{-1} + e}{2} = \cosh 1 \\ sin x = e i x e i x 2 i sin i = e i 2 e i 2 2 i = e 1 e 2 i = e e 1 2 i = i sinh 1 \sin x = \dfrac {e^{ix} - e^{-ix}}{2i} \Rightarrow \sin i = \dfrac {e^{i^2} - e^{-i^2}}{2i} = \dfrac {e^{-1} - e}{2i} = -\dfrac {e - e^{-1}}{2i} = i\sinh 1 \\


cos ( 1 + i ) = cos 1 cosh 1 i sin 1 sinh 1 \therefore \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 = 0.834 \Rightarrow a = Real Part = \cos 1\cosh 1 = \boxed{0.834}


Hint \rightarrow Angles in radians .

Forgot to put angle in radians and got all three tries wrong :p

Keshav Gupta - 5 years, 6 months ago

Why the 1 in cos 1 have to be in radians?

AccelNano Lim Loong - 6 years, 3 months ago

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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.

Sudeshna Pontula - 6 years, 3 months ago

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Thanks for pointing out

AccelNano Lim Loong - 6 years, 3 months ago

We can write c o s ( x ) = e i x + e i x 2 cos(x) = \frac{e^{ix} + e^{-ix}}{2} , derived from Euler's formula

c o s ( 1 + i ) = e 1 + i + e 1 i 2 cos(1 + i) = \frac{e^{-1 + i} + e^{1 - i}}{2}

Let y = e 1 + i = e 1 e i 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 ) e^{ix} = cos(x) + isin(x) ; thus, e i = c o s ( 1 ) + i s i n ( 1 ) e^{i} = cos(1) + i*sin(1) ; therefore, y = c o s ( 1 ) + i s i n ( 1 ) e y = \frac{cos(1) + i*sin(1)}{e} , and from this we conclude that c o s ( 1 + i ) = y + 1 y 2 = 0.833730025 0.988897706 i cos (1 + i) = \frac{y + \frac{1}{y}}{2} = 0.833730025 - 0.988897706i .

The problem asks for the real part of c o s ( 1 + i ) cos(1 + i) , which equals approximately 0.833 0.833 .

Chenyang Sun
Mar 30, 2015

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!

Akeel Howell - 4 years ago
Jeremy Weissmann
Nov 24, 2016

I used the series

cos ( z ) = n ( 1 ) n z 2 n / ( 2 n ) ! \cos(z) = \sum_n (-1)^n \cdot z^{2n}/(2n)!

Letting z = 1 + i = 2 e i π / 4 z = 1+i = \sqrt{2} \cdot e^{i \cdot \pi/4} \ , we have

cos ( z ) = n ( 1 ) n 2 n e i n π / 2 / ( 2 n ) ! \cos(z) = \sum_n (-1)^n \cdot 2^n \cdot e^{i \cdot n\pi/2}/(2n)!

Now,

e i n π / 2 = { ( 1 ) m i f $ n = 2 m $ ( 1 ) m i i f $ n = 2 m + 1 $ e^{i \cdot n\pi/2} = \cases{(-1)^m & if \$\ n = 2m\$ \cr (-1)^m \cdot i & if \$\ n = 2m+1\$ \cr }

and therefore:

R e cos ( z ) = m ( 1 ) m 4 m / ( 4 m ) ! = 1 4 4 ! + 16 8 ! + 0.834 {\rm Re} \cos(z) = \sum_m (-1)^m \cdot 4^m / (4m)! = 1 - {4 \over 4!} + {16 \over 8!} + \cdots \approx 0.834

The Alternating Series Error Rule tells us that two terms will suffice (if both truncating and rounding are allowed), since 16 8 ! < 0.0005. {16 \over 8!} < 0.0005.

Can we use Taylor Series here???I did the same and got the answer.

Definitely! TS apply equally in R and C. 😉

Jean-Baptiste Pomarède - 4 years, 10 months ago
Karthick Shiva
Feb 11, 2016

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..

Sagar Shah - 5 years ago

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what about graphing r=cos(i+1), am i booted out of the club

Akira Kato - 4 years, 8 months ago

Thts unethical on brilliant buddy 😁

will jain - 4 years, 7 months ago

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