Author Topic: solving (Read 1196 times) Tweet Share

sportzstar77 · « **on:** April 06, 2012, 04:23:09 pm »

solve with method please: z^2-(3+4i)=0 CHEERS

answer: 2+i, -2-i

leejieun94 · « **Reply #1 on:** April 06, 2012, 04:45:07 pm »

let z=x+yi and go from there

pi · « **Reply #2 on:** April 06, 2012, 04:50:31 pm »

Hint: factorise it

tony3272 · « **Reply #3 on:** April 06, 2012, 04:56:13 pm »

$z^2-3+4i=0$

$\implies (x+yi)^2-3+4i=0$

$\implies (x^2-y^2-3)+(2xyi+4i)=0$

$\implies 2xyi=-4i$ and $x^2-y^2-3=0$

From the first one:
$y=\frac{-2}{x}$

$\implies x^2-\frac{4}{x^2}-3=0$

$\implies x^4-3x^2-4=0$

$\implies x^2=4$ or $x^2=-1$ However we ignore the negative solution as x E R

$\implies x=\pm 2$

$\implies y=\mp 1$

$\implies z=-2+i$ and $z=2-i$

Bhootnike · « **Reply #4 on:** April 06, 2012, 11:11:11 pm »

Quote from: tony3272 on April 06, 2012, 04:56:13 pm

$z^2-3+4i=0$

$\implies (x+yi)^2-3+4i=0$

$\implies (x^2-y^2-3)+(2xyi+4i)=0$

$\implies 2xyi=-4i$ and $x^2-y^2-3=0$

From the first one:
$y=\frac{-2}{x}$

$\implies x^2-\frac{4}{x^2}-3=0$

$\implies x^4-3x^2-4=0$

$\implies x^2=4$ or $x^2=-1$ However we ignore the negative solution as x E R

$\implies x=\pm 2$

$\implies y=\mp 1$

$\implies z=-2+i$ and $z=2-i$

for OP, you may wonder why answer isnt the same as your answers give, but dw, follow same method , its just that tony misread the original equation

shouldve been $z^2-3-4i=0$ , -4i not 4i

tony3272 · « **Reply #5 on:** April 06, 2012, 11:27:25 pm »

Haha my bad... totally forgot the brackets!

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Author Topic: solving (Read 1196 times) Tweet Share

sportzstar77

solving

leejieun94

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pi

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tony3272

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Bhootnike

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tony3272

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