Nice #3.

"I always make mistakes " 😂😂😂 I spotted the a^2 error and started writing thus message even I saw you correct it 😂

z=1 is an obvious solution Bring everything to the lhs: z^2 + (i-1)z -1=0 divide lhs (using long division) by z-1, since z=1 is a solution you get z+i+1, so you can factor z^2 + (i-1)z -1 = (z-1)(z+i+1) = 0 This gives z=1 or z=-1-i

Got 'em both!

I used the second method.

You had a product of 2 quadratics which you have set to zero, which means you could've used the quadratic formula. 4th method? Writing z and the righthand side in polar form.

A mistake@5:57 it equals to 1 not 0

This is how I like solving this kind of equation: |z|-4=2i-iz, so that 2i-iz must be a real number, say t. Then 2i-iz=t -> z=2+it and substituting in the initial equation we get sqrt(t²+4)=4+t and we conclude as in the video

But this is complex root and it has 2 roots?

Hello can anyone please tell me what kind of math is this. Trig calculus, physics 🤷

Ans you need ro write down thwt z^2 can't be zero,otherwise we would dividd by zero

It's easier if you make z^2= a,solve and then substitute back

I started down another method - multiply by the conjugates which gives |z|^2 + |z|(z - /z)i - |z|^2 = 4^2 + 2^2 2b|z| = 20 But then I watched your video and your method is easier 😂

Got it!

Z= -3i -1

problem |z| + i z = 4 + 2 i Let z = a + b i |z| = √(a² + b²) √(a² + b²) + i a - b = 4 + 2 i √(a² + b²) - b = 4 a = 2 √(4 + b²) = 4 + b Square. 4 + b² = b² + 8 b + 16 8 b = -12 b = -3/2 answer z = 2 - 3/2 i

Nice problem that i solved in my head😊

(sqrt(5) +/- 1)/2

You really should have finished your first method. At 5:53 we have (z² + ¹⁄₂z + k)² = (2k − ³⁄₄)z² + (k − 1)z + (k² − 1) Do you see what I see? People shouldn't believe you when you say that solving this is going to be super time consuming, because it is not. Both the coefficient of the linear term _and_ the constant term of the quadratic in z at the right hand side reduce to zero for k = 1 (making its discriminant equal to zero) and this gives us (z² + ¹⁄₂z + 1)² = ⁵⁄₄z² or (z² + ¹⁄₂z + 1)² = (¹⁄₂√5·z)² which is easy to solve.

problem z⁵ = 1 Use the polar form of 1. 1 = e^(i 2πN), N ∈ ℤ z⁵ = e^(i 2πN) Take to (1/5) power. z = e^(i 2πN/5) The distinct values of N are from 0 to 4, then the angles repeat. N=0, angle 0° z = 1 N=1, angle 72° z = e^(i 2π/5) = (√5-1)/4 + i√(10+2√5)/4 N=2, angle 144° z = e^(i 2πN/5) = -(√5+1)/4 + i √(10-2√5)/4 N=3, angle 216° z = e^(i 2πN/5) = -(√5+1)/4 - i √(10-2√5)/4 N=4, angle 288° z = e^(i 2πN/5) = (√5-1)/4 - i√(10+2√5)/4 answer z ∈ { 1, (√5 - 1) / 4 + i √(10+2√5)/4, -(√5 + 1) / 4 + i √(10-2√5)/4, -(√5 + 1) / 4 - i √(10-2√5)/4, (√5 - 1) / 4 - i √(10+2√5)/4 }

There's a neat solution for finding cos(2pi/5). Express in terms of cos(pi/5) and then equate cos(4pi/5) to cos(pi - pi/5). All using double angle formulae and a quadratic equation.

Why this channel couldn't be (Re) named ? R*e(i*arg(z) ) for example ? 😉 When polynome degree is high, polar form is often more efficient.

This is just a 5th roots of unity problem; thus the solutions are e^(0ipi/5), e^(2ipi/5), e^(4ipi/5), e^(6ipi/5), and e^(8ipi/5).

Nice!

z√z=2√2+2√2i =4(1/√2+i/√2) =4e^(iπ/4) ∴z=³√16·e(iπ/6) =³√16(√3/2+1/2·i) =³√2(√3+i)

problem z√z = 2√2 + 2√2 i z^(3/2) = 2√2 ( 1 + i ) Use the polar form of 1 + i. 1 + i = √2 e^( i π (1/4 + 2N) , where N is an integer. z^(3/2) = 2√2 [√2 e^( i π (1/4 + 2N) ] z^(3/2) = 4 [ e^( i π (1/4 + 2N) ] Raise both sides to the power 2/3. [z^(3/2)]^(2/3) = [(4^(2/3)] [ e^( 2 i π (1/4 + 2N) /3] z = (2∛2) e^[ i π ( 8N + 1 ) /6 ] , N = 0,1,2 At N=3 the angles start to repeat. Thus the set of distinct answers is z = (2∛2) ( cos π /6 + i sin π /6) = (2∛2) (√3/2 + i / 2) z = (2∛2) ( - i ) z = (2∛2) (-√3/2 + i / 2) answer z ∈ { -2 ∛2 i, ∛2 (-√3+ i ), ∛2 ( √3 + i ) }

You were doing well till the end!!! n = 0, 1 or 2.😂 Of course the "doing it in my head" version was noticing that the rhs has a modulus of 4 and an argument of pi/4. So z^3 has modulus of 16 and argument of pi/2. z = cbrt(16) * [cube roots of i]. But will there be extraneous solutions because of the sqrt sign in the original equation??? That's something to think about... 😉

Nice!

why can't u just wait until the END to think about minor roots, i.e,- z^3 = 16i = 16e^(PI/2), z = 16^(1/3) * (e^ i [PI/2]) ^ (1/3) = 16^(1/3) * e^ i [PI/6]. Now we add 2nP: z = 16^(1/3) * e^ i [ (PI/6) + 2nPI] ?

z^(3/2) = 2sqrt(2)*(1+i) z^(3/2) = 4*[1/sqrt(2) + i/sqrt(2)] <-- this is now a unit vector z^(3/2) = 4*e^(i*pi/4) <-- also, plus 2pi*n, if you want all the solutions here z = [4*e^(i*pi/4)]^(2/3) z = cbrt(4^2)*e^(i*pi/6) z = 2*cbrt(2)*[sqrt(3)/2 + i/2] z = cbrt(2)*sqrt(3) + cbrt(2)*i

All good comments, folks!

problem z ⁱ = e ⷫᐟ² Take both sides to the power -i. ( z ⁱ )⁻ⁱ = ( e ⷫᐟ² )⁻ⁱ Multiply exponents. z = e⁻ⁱ ⷫᐟ² This is only the principle root. Add n times 2π to the angle for the general case. z = e⁻ⁱ ⷫᐟ² = e ⁻ⁱ ⷫ ⁽¹ᐟ²⁺²ⁿ⁾, n ∈ ℤ Since n is an integer however, this always comes back to z = -i answer z = - i

Raise both sides to the power i and you can miss out the ln bits.

Your final solution, z = e^{-i(π/2 + 2πn)} is just z = -i. But there are actually infinitely many solutions of which this is only one. Solution is z = (-i)e^(2kπ) ∀k∈Z.

I would start with z = r *(e^(i*(theta))). So now we got e on both sides.

Just draw e^(-i pi/2) in the complex plane. r = 1 theta = - pi/2 => (-i)

I also got -i.

How about a shorter path: raise both sides to the power -i and there's your principal solution. because (z^i)^-i = z^1 Slightly longer is raising both sides to 1/i, but that's the same

The equation only involves the reciprocal of z. So multiplying by 25 and defining w=25/z will make things much easier. In terms of w, the equation reads w + |w| = 9 + 3i ** |w| is real, so we read off that the imaginary part of w must be 3: w = c + 3i. Equating the real parts of **, we then get c + sqrt(c^2 + 3^2) = 9. This is not hard to solve and results in c=4. We have w, and can take a reciprocal to get back to z. 25/z = w = 4 + 3i z = 25/(4 + 3i) = 4 - 3i.

Can we not go directly from I^z=ni The very end Zlni= ln(n) +lni z= ln(n)/ln(i) +1 z= 1+ ln(n)/( ia π )) z=1 -i ln(n)/(a.π) where a= 2k+1/2

ni?! Be very careful, the knights may be lurking

problem i ᶻ = n i Assume n is integer. Take the natural logarithm of each side. Bring down exponents. z ln i = ln n + ln i Subtract ln i from each side. z ln i - ln i = ln n Factor. z ln i - ln i = ln n ( z - 1 ) ln i = ln n Substitute for ln i. ln i = i π ( 1/2 + 2 K ), K an integer. ( z - 1 ) i π ( 1/2 + 2 K ) = ln n Multiply by -i. ( z - 1 ) π ( 1/2 + 2 K ) = - i ln n Divide by π ( 1/2 + 2 K ). ( z - 1 ) = - i ln n / [ π ( 1/2 + 2 K ) ] z = 1 - 2 i ln n / [ π ( 4 K + 1 ) ] answer z ∈ { 1 - 2 i ln n / [ π ( 4 K + 1 ) ], ( n, K ∈ ℤ ) }

At the end, you used k on both sides. Then cancelled. This is wrong. Mostly. You should have not 1 but (1 + 4k)/(1 + 4m)

Cool!

Use Euler formula

problem 1/z + 1 / |z| = ( 9 + 3 i ) / 25 z • z̅ = | z |² , where z̅ is the complex conjugate of z. 1/ |z| = 1 / √(z • z̅ ) = √(z • z̅ ) / (z • z̅ ) 1/z + √(z•z̅ ) / (z•z̅ ) = ( 9 + 3 i ) / 25 [ z̅ +√(z•z̅ ) ] / (z•z̅ ) = ( 9 + 3 i ) / 25 Replace z with a+ b i. z = a + bi [ a - bi +√(a²+ b² ) ] / (a²+ b²) = ( 9 + 3 i ) / 25 (a²+ b²) = 25 √(a²+ b² ) = 5 ( a - bi + 5 ) / 25 = ( 9 + 3 i ) / 25 a - bi + 5 = 9 + 3 i a - bi = 4 + 3 i a = 4 b = -3 answer z = 4 - 3 i

Let the RHS be C. Isolate the z-term to get 1/z = C - 1/|z|. Take the absolute value of both sides and square the results to get 1/|z|^2 = |C|^2 - (C+Cbar)/|z| + 1/|z|^2. Then |C|^2 - (C+Cbar)/|z| = 0. Solve for |z| to get |z| = 5. Use this result in the first equation to get zbar = 4 + 3 i so that z = 4 - 3 i.

Thank you. Here is an alternative way. ( _Using _*_j = √(-1)_*_ instead of _*_i_*_ because I can't get a suitable _*_i_*_ superscript._ ) *_1/z + 1/|z| = (9 + 3j)/25_* ... ① ⇒ _(|z| + z)/z|z| = (9 + 3j)/25_ ⇒ _(1 + z/|z|)/z = (9 + 3j)/25_ Let *_z = reʲᵗ = r( cos(t) + jsin(t) )_* where _r = |z|_ and _t_ is some angle. ∴ _(1 + eʲᵗ)/reʲᵗ = (9 + 3j)/25_ ⇒ _r⁻¹(1 + e⁻ʲᵗ) = (9 +3j)/25_ ⇒ _1 + cos(-t) + jsin(-t) = (9r + 3rj)/25_ ⇒ _cos(t) - jsin(t) = (9r - 25)/25 + 3rj)/25_ ⇒ *_sin(t) = -3r/25, cos(t) = (9r - 25)/25_* ... ② ∴ _1 = sin²(t) + cos²(t) = (9r² + 81r² - 450r + 625)/625_ ⇒ _90 r² - 450r + 625 = 625_ ⇒ *_r(r - 5) = 0_* _r = 0 ⇒ |z| = 0_ which does not work in ① ∴ _r = 5_ ∴ from ②: _sin(t) = -⅗, cos(t) = ⅘_ ∴ *_z = 5(⅘ - ⅗j) = 4 - 3j_*

Nice!

a = kb and b=−√b^2 since b < 0. k−√(k^2+1) = −3 and k = −4/3.