An Absolutely Nice Homemade Equation | Problem 473

[@angelishify]

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

[@scottleung9587]

Nice!

[@siavashrahimian1514]

Use Euler formula

[@mcwulf25]

Bit of a guess at the end there. I wonder if it's a case of substituting a^2 + b^2 = 25c^2 ?

[@matthewfeig5624]

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.

[@lhdill2009]

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.

[@Don-Ensley]

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

[@guyhoghton399]

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_*

[@mab9316]

kzbin.info/www/bejne/roTUlHewe8aYe7s

[@actions-speak]

Let w = 1/z and a + bi = w/|w|. Then (9+3i)/25 = w + |w| = |w| (a + bi + 1) and a + bi = (9 - |w|25 + 3i)/(|w|25) where |a + bi| = 1, so |a + bi|^2 = 1 and (9 - |w|25 + 3i)(9 - |w|25 - 3i)/(|w|25)^2 = 1, and after a little algebra, |w| = 1/5, a + bi = (4 + 3i)/5, w = (4 + 3i)/25, z = 25 (4 - 3i)/(4^2 + 3^2) = 4 - 3i.