Elementary Complex Number

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

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

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

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

$\dfrac{1}{2}(\cos(x)(e^{y} + e^{-y}) - i\sin(x)(e^{y} - e^{-y})) = \cos(x)\cosh(y) - i*\sin(x)\sinh(y)$ ,

where the definitions $\cosh(y) = \dfrac{e^{x} + e^{-x}}{2}$ and $\sinh(y) = \dfrac{e^{y} - e^{-y}}{2}$ were used.

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

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

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

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

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Hint $\rightarrow$ Angles in radians .

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

$cos(1 + i) = \frac{e^{-1 + i} + e^{1 - i}}{2}$

Let $y = e^{-1 + i} = e^{-1}*e^{i}$

Now, Euler's formula states that $e^{ix} = cos(x) + isin(x)$ ; thus, $e^{i} = cos(1) + i*sin(1)$ ; therefore, $y = \frac{cos(1) + i*sin(1)}{e}$ , and from this we conclude that $cos (1 + i) = \frac{y + \frac{1}{y}}{2} = 0.833730025 - 0.988897706i$ .

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

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.

I used the series

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

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

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

Now,

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

and therefore:

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

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

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

Just google,

cos(1+i)

;) - simple