The problem is the cosine function works with real numbers. If you want it to output imaginary numbers, you have to start with an imaginary cosine function. Once you extend it into the complex number field, it is no longer the same function.

Given that cosine can be 2 with the domain expanded to complex numbers, what values of Cos x can never exist?

😎👍👏👍😎

Cool!

I am the new Euler. I got the answer in 5 seconds. The answer is arccos(2) 👋😁

problem cos θ = 2 Preliminary note: By even symmetry of the cosine, for any solution θ found, -θ is also a solution. Use the complex cosine definition based on Euler's formula. cos θ = ( e ⁱ ᶿ + e ⁻ ⁱ ᶿ ) / 2 Set this equal to 2. ( e ⁱ ᶿ + e ⁻ ⁱ ᶿ ) / 2 = 2 Multiply by 2. e ⁱ ᶿ + e ⁻ ⁱ ᶿ = 4 Let y = e ⁱ ᶿ y + 1 / y = 4 y² - 4 y + 1 = 0 ( y - 2 )²- 3 = 0 ( y - 2 )² = 3 y - 2 = ± √3 y = 2 ± √3 Back substitute. e ⁱ ᶿ = y e ⁱ ᶿ = 2 ± √3 This is positive for either sign of the radical. Take natural logarithms. i θ = ln ( 2 ± √3 ) Multiply both sides by -i. - i • i θ = -i ln ( 2 ± √3 ) - i² θ = -i ln ( 2 ± √3 ) - (-1) θ = -i ln ( 2 ± √3 ) θ = -i ln ( 2 ± √3 ) and by symmetry, θ = i ln ( 2 ± √3 ) Interestingly, because (2+√3) = (2+√3)(2-√ 3)/(2-√3) = 1/(2-√3) then ln ( 2+√3 ) = - ln ( 2-√3 ) So -i ln ( 2 + √3 ) = i ln ( 2 - √3 ) and -i ln ( 2 - √3 ) = i ln ( 2 + √3 ) So the symmetry consideration is in this case redundant. Because cosine has a period of 2π, then if θ is a solution then so is θ + 2π N where N is an integer. θ = 2πΝ + i ln ( 2 ± √3 ) , where N ∈ ℤ answer θ ∈ { 2πΝ + i ln ( 2 - √3 ), 2πΝ + i ln ( 2 + √3 ), ( N ∈ ℤ ) }

This series of lectures should be enough to provide a basis for a course in complex analysis. Thank you for providing this!😎

Didn't you do this problem already? Maybe I'm thinking of another math youtuber.

I agree with the solution. Curiously WA, give for principal value (n=0) x= arcos(2) or x= - arcos (2) , without further development .

I came up with: z_1(n) = ((3π/4)+2πn) + i(-ln((sqrt(3)+1)/sqrt(2))) z_2(n) = ((7π/4)+2πn) + i(-ln((sqrt(3)-1)/sqrt(2)))

Nice vid! Another quicker way is as we know that cos(theta) = 2, and sin(theta) = ± (√3) i, then by Euler's formula e^i(theta) = cos(theta) + i sin(theta) = 2 + i (±√3 i) = 2 ± √3, and then we do the same as you and turn it to a complex number.

Bsquare root of both sides and used twice will be the result 😅 Z = 2

I don’t think you need to make trigonometry videos. Unlike complex analysis, knowledge of trigonometry is a prerequisite for calculus, which is what you are doing. I also noticed you didn’t advertise that you were @sybermath in this video. Did you forget?

Hello my friend. I just found this extra channel, so I'm subscribing to it. I have to keep up with your latest venture!😎

problem ln(z+i) = 1+i z + i = e^(1+i) z +i = e^(1+i) = e¹ • e^i = e • e ⁱ = e ( cos 1 + i sin 1 ) z = e cos 1 + i ( e sin 1 - 1 ) answer z = e cos 1 + i ( e sin 1 - 1 )

Z= e^(1+i) -i

(24)^2 (12i)^2={576 ➖ 144i^2}=432i 10^40^32 10^4^10^4^8 2^5^2^2^2^5^2^2^2^3 1^1^1^1^1^1^1^1^1^1^2 1^2 (z➖ 2z+1i).

Like others I also used the second method, but I find the first one way more interesting. Well done!

I couldn't see the answer to the question, its resolution is very confusing. Say the value Z = ?

I used the second method.

I immediately knew that you could exponentiate the ln

Plus 2.pi.n.i

How did you get to the next term

problem z̅ + z |z| = 24 - 12 i Define z = x + i y z̅ + z |z| = 24 -12 i Translates to x - i y + ( x + i y) √( x² + y² ) = 24 - 12 i Let √( x² + y² ) be denoted by m for modulus. x - i y + ( x + i y) m = 24 - 12 i x ( m + 1 ) + i y ( m - 1 ) = 24 - 12 i Separating out real and imaginary parts' equations x ( m + 1 ) = 24 y ( m - 1 ) = -12 x² = [24/(m+1) ]² = (24²)/(m² +2m+1) y² = [12/(m-1) ]² = (12²)/(m²-2m +1) x² +y² = m² =[ (24²) ( m-1)² +(12²) (m+1)²] / [(m²-1)² ] Expand. m⁶-2 m⁴-719 m² + 864 m -720 = 0 m = 5 is a root by RRT. This is the only positive non-complex root which we need for the modulus. x ( 5 + 1 ) = 24 y ( 5 - 1 ) = -12 6x = 24 4y = -12 x = 4 y = -3 answer z = 4 - 3i

The curve I found (not that it is correct) has the shape in polar coordinates of a sideways teardrop extending to infinity every 2π radians due to the ln (sin θ/2) term. problem Can you sum 1 + cos θ / 2 + cos 2θ / 4 + cos 3 θ / 6 + ... Call it S. In sigma notation, with leading 1 taken separately, the sum, S, is   ͚ S = 1 + Σ cos nx /( 2 n )   ⁿ⁼¹ Recall Euler's formula e ⁱ ᶿ = cos θ + i sin θ Re(e ⁱ ᶿ) = cos θ Recall the series representation for ln(1-x). ͚ ln(1-x) = - Σ ( xⁿ ) / n   ⁿ⁼¹ Let     ͚ P = Σ cos nx /( 2 n )  ⁿ⁼¹ S = 1 + P     ͚ P = ½ Σ cos nx / n   ⁿ⁼¹ ͚ Re( ln(1-e ⁱ ᶿ) )= - Σ ( Re(e ⁱ ᶿ ⁿ) ) / n   ⁿ⁼¹ , which is in terms of P Re( ln(1-e ⁱ ᶿ) )= -2 P Solve for P. P = - ½ Re( ln(1-e ⁱ ᶿ) ) S = 1 - ½ Re( ln(1-e ⁱ ᶿ) ) All now to do is evaluate Re( ln(1-e ⁱ ᶿ) ) = Re ( ln ( 1 - cos θ - i sin θ ) ) Focusing on the argument of ln, 1 - cos θ - i sin θ Using double angle formulas, cos θ = cos [2 (½θ)] = cos²( θ/2) - sin²( θ/2) = 1- 2 sin²( θ/2) sin θ = sin [2 (½θ)] = 2 sin (θ/2) cos(θ/2) 1 - cos θ - i sin θ = 1-(1- 2 sin²( θ/2))- i 2 sin (θ/2) cos(θ/2) = 2 sin²( θ/2) - i 2 sin (θ/2) cos(θ/2) = 2 sin (θ/2) [ sin (θ/2) - i cos(θ/2) ] = 2 sin (θ/2) [ sin (θ/2) - i cos(θ/2) ] = -2 i sin (θ/2) [ cos(θ/2) + i sin (θ/2) ] = -2 i sin (θ/2) [ e ⁱ ᶿᐟ² ] So the argument to ln is -2 i sin (θ/2) [ e ⁱ ᶿᐟ² ]. ln ( 1 - cos θ - i sin θ ) = ln { -2 i sin (θ/2) [ e ⁱ ᶿᐟ² ]} = ln ( -2 i sin (θ/2)) + i θ / 2 = ln(-2i) + ln[ sin (θ/2) ]+ i θ / 2 = -2i has modulus 2 and angle -π/2. Polar form of -2i is -2i = 2 e^(-i π/2) ln ( 1 - cos θ - i sin θ ) = ln(2 e^(-i π/2)) + ln[ sin (θ/2) ]+ i θ / 2 = ln 2 -i π/2 + ln[ sin (θ/2) ]+ i θ / 2 = ln 2 + ln[ sin (θ/2) ]+ i (θ-π) / 2 Re( ln(1-e ⁱ ᶿ) ) = ln 2 + ln[ sin (θ/2) ] S = 1 - ½ { ln 2 + ln[ sin (θ/2) ] } = 1 - ½ ln 2 - ½ ln [ sin (θ/2) ] answer 1 - ½ ln 2 - ½ ln [ sin (θ/2) ]

I also got 4-3i as the answer - good job, WA!

(0.2Sqrt[5+240Sqrt[5]])-0.2Sqrt[5]+i(0.1Sqrt[5]-0.1Sqrt[5+240Sqrt[5]])+(0.2Sqrt[5+240Sqrt[5]])-0.2Sqrt[5]+i(0.1Sqrt[5]-0.1Sqrt[5+240Sqrt[5]])Abs[(0.2Sqrt[5+240Sqrt[5]])-0.2Sqrt[5]+i(0.1Sqrt[5]-0.1Sqrt[5+240Sqrt[5]])]=24-12i -1/Sqrt[5]+Sqrt((1/5)(1 + 48Sqrt(5)))+i((1/2)Sqrt((1/5)(1 + 48 sqrt(5)))-1/(2Sqrt[5]))+-1/Sqrt[5]+Sqrt((1/5)(1 + 48Sqrt(5)))+i((1/2)Sqrt(1/5 (1 + 48 sqrt(5)))-1/(2Sqrt[5]))+Abs[ -1/Sqrt[5]+Sqrt((1/5)(1 + 48Sqrt(5)))+i((1/2)Sqrt(1/5 (1 + 48 sqrt(5)))-1/(2Sqrt[5]))]=24-12i Z=(0.2Sqrt[5+240Sqrt[5]])-0.2Sqrt[5]+i(0.1Sqrt[5]-0.1Sqrt[5+240Sqrt[5]])

Very ez

(1+i)/2

problem zᵉ^ᶻ = -i / π Raise both sides to power (1/z). [ zᵉ^ᶻ ]⁽¹ᐟᶻ⁾ = (-i / π) ⁽¹ᐟᶻ⁾ zᵉ = (-i / π) ⁽¹ᐟᶻ⁾ Take lns. Bring down exponents. e ln z = (1/z) ln (-i / π ) Polar form of -i is -i = e^[ -iπ(½+2N)] , where N is an integer. Divide by e and multiply by z. z ln z = (1/e) [ - ln π - iπ(½+2N) ] ln z e ˡⁿ ᶻ = (1/e) [ -ln π - iπ(½+2N) ] ln z = W [- (ln π)/e - iπ (½+2N)/e ] z = e^ { W [- (ln π)/e - iπ (½+2N)/e ] } answer z = e^ { W [- (ln π)/e - iπ (½+2N)/e ] }, ( N ∈ ℤ )

How did I end up here.

Another Wolfram solution was 2 + 4i. Not seeing that!

You made the rhs hard work. -i/pi = -1/i.pi doesn't require any ln stuff.

NO WAY YOURE SYBERMATH IM SUBBED TO YOU BOTH OMG YAY

2:37 you could give it SUM_SYMBOL (capital sigma) n=1 to infinity of (i^n)/n

Nice😊

Nice!

same answer with your answer

Method 2. we will use nth powers of i here. i +i² /2 +i³ /3 +i^4 /4 +... = i -1 /2 -i /3 +1 /4 +... Let's separate the last statement into real and imaginary parts. =(-1 /2 +1 /4 -1 /6 +...) +(1 -1 /3 +1 /5 -1 /7 +...)×i. Sum the series in the real part. -1 /2 +1 /4 -1 /6 +... we know from series theory that 1 -1 /2 +1 /3 -1 /4 +... =ln2. then implies that =-1 /2 ×(1 -1 /2 +1 /3 -1 /4 +...) =-ln2 /2. now Sum the series in the imaginary part. we know that 1 -1 /3 +1 /5 -1 /7 +... =π /4. and our final answer is -ln2 /2 +iπ /4.

x+x^2/2+x^3/3 + .... = -ln(1-x). Set x = i. So, S=-ln(1-i) = -ln[√2 e^(7pi i/4)] = -1/2 ln 2 -7 pi i/4.

I love your videos, I've been binge watching them and I hope this albeit small amount of money helps. I'm low income so can't do much more :(

(-i)^2+(-i)^3+(-i)^4 = -1+i+1 = i

With the sound being about 5 seconds out of synch with the picture, this is almost unwatchable.

First, consider this: i¹=i i²=-1 i³=-i i⁴=1 Now we can solve the problem. S= i+i²/2+i³/3+i⁴/4+...+iⁿ/n+... S= i-(1/2)-(i/3)+(1/4)+...+iⁿ/n+... S= iΣ[1/(4n+1)] -Σ[1/(4n+2)] -iΣ[1/(4n+3)] +Σ[1/(4n+4)] all this sums are finite because the terms lie between 0 and 1 Make the calculus to obtain each sum, operate the anwsers and done.

8:58 mmm yes remove absolute value.

problem Can you sum i + i² + i³ + i⁴ + ... Call the sum S. Using sigma notation ∞ S = Σ (i ⁿ) / n ⁿ⁼¹ Suppose f(x) is defined as ∞ f(x) = Σ xⁿ/n ⁿ⁼¹ Take the derivative of f with respect to x. ∞ f'(x) = Σ nx⁽ⁿ⁻¹⁾/n ⁿ⁼¹ ∞ f'(x) = Σ x⁽ⁿ⁻¹⁾ ⁿ⁼¹ , which with shift of index becomes ∞ f'(x) = Σ x ⁿ ⁿ⁼⁰ The first term is 1 and the ratio between consecutive terms is x. The sum s of this geometric series is found as s = a / ( 1 - r ) = 1 / ( 1 - x ) This is the derivative of the desired function. Its antiderivative is f(x) = ∫ dx / ( 1 - x ) f(x) = - ln(1-x) By knowing ∞ Σ xⁿ/n = - ln(1-x) ⁿ⁼¹ we can replace x with i. 1-i in polar form using the Argand plane is (1-i) = √2 e⁻ⁱ ⷫᐟ⁴ The replacement of x by i gives us ∞ S = Σ iⁿ/n = - ln(1-i) ⁿ⁼¹ = - ln√2 + i π /4 = -½ ln 2 + ¼ i π answer S = -½ ln 2 + ¼ i π

{i+i ➖ }+{i^2+i^2 ➖ }/{2i+2i ➖ }={i^2+i^4}/4i^2=i^6/4i^2+{i^3+i^3 ➖ }{3i+3i ➖ }={i^6/4i^2+i^6/6i^2}=i^12/10i^4+{i^4+i^4 ➖ }/{4i+4i ➖}={i^11/10i^4+i^8/8i^2 }=i^19/18i^6 +{i^5+i^5 ➖ }/{5i+5i ➖ }={i^19/18i^6+i^10/10i^2}=i^29/28i^8+{i^6+i^6 ➖ }/{6i+6In ➖ }={i^29/38i^8+i^12/12i^2}=i^^41/50i^10 i^12^29/12^38i^8 i^12^13^16/12^19^19i^8 i^6^6^13^8^8/6^6^9^10i^8 i^2^3^2^36^7^4^4/2^3^2^3^9^5^5i^4^4 i^1^1^1^1^6^3^4^2^2^2^2/1^1^1^13^2^2^3^2^3i^1^1 i^2^3^3^2^2^1^1^1^1/1^1^1^1^1^1i i^1^1^3^1^2/ i/3^2/ (i ➖ 3pi+2). I HAVE NEVER SEEN IMAGINARY INFINITY SUM BEFOR THAT IS INTERSET EDUCATIONAL INFORMATIVE VIDEO THANK YOU.

Man, your videos are amazing!!! Although I have noticed that there has been some video delay with respect to the audio, it'd be cool if you fixed that

i pi/4 - 1/2 ln 2 Nice calculation! Or you remember that 1 - 1/3 + 1/5 - 1/7 +... = pi/4 and 1 - 1/2 + 1/3 - 1/4 +... = ln 2 from Analysis I, the first with factor i, the second with 1/2 i^2 = -1/2.

Your error with your first method is that you took arctan(b/a) to be π, this is incorrect. The arctangent of a real number is always on the open interval (−½π, ½π) whereas the principal argument of a nonzero complex number can have any value on the interval (−π, π]. This implies that arctan(b/a) only gives the principal argument for a complex number a + bi in the _right half_ of the complex plane.