Solve-the-equation-z-2-2-1-j-z-2-0-Givin-each-result-in-form-a-jb-with-a-and-b-correct-to-2dp-

Question Number 15348 by tawa tawa last updated on 09/Jun/17

Solve the equation z^{2} + 2 (1 + j) z + 2 = 0, Givin each result in form a + jb

with a and b correct to 2 dp .

Answered by RasheedSoomro last updated on 10/Jun/17

Let z = x + jy

z^{2} + 2 (1 + j) z + 2 = 0

{(x + jy)}^{2} + 2 (1 + j) (x + jy) + 2 = 0

x^{2} - y^{2} + 2 xyj + 2 (x + xj + yj - y) + 2 = 0

x^{2} - y^{2} + 2 xyj + 2 x + 2 xj + 2 yj - 2 y + 2 = 0

(x^{2} - y^{2} + 2 x - 2 y + 2) + (2 xy + 2 x + 2 y) j = 0 + 0 j

x^{2} - y^{2} + 2 x - 2 y + 2 = 0 \land 2 xy + 2 x + 2 y = 0

(x - y) (x + y) + 2 (x - y) + 2 = 0 \land x + xy + y = 0

(x - y) (x + y + 2) + 2 = 0 \land y = \frac{- x}{x + 1}

(x - \frac{- x}{x + 1}) (x + \frac{- x}{x + 1} + 2) + 2 = 0

\frac{x^{2} + x + x}{x + 1} \times \frac{x^{2} + x - x + 2 x + 2}{x + 1} + 2 = 0

\frac{x^{2} + 2 x}{x + 1} \times \frac{x^{2} + 2 x + 2}{x + 1} + 2 = 0

{(x^{2} + 2 x)}^{2} + 2 (x^{2} + 2 x) + 2 (x^{2} + 2 x + 1) = 0

Let x^{2} + 2 x = t

t^{2} + 2 t + 2 t + 2 = 0

t^{2} + 4 t + 2 = 0

t = \frac{- 4 \pm \sqrt{16 - 8}}{2}

t = - 2 \pm \sqrt{2}

x^{2} + 2 x = - 2 + \sqrt{2} \lor x^{2} + 2 x = - 2 - \sqrt{2}

x^{2} + 2 x + 2 - \sqrt{2} = 0 \lor x^{2} + 2 x + 2 + \sqrt{2} = 0

x = \frac{- 2 \pm \sqrt{4 - 4 (2 - \sqrt{2})}}{2} \lor x = \frac{- 2 \pm \sqrt{4 - 4 (2 + \sqrt{2})}}{2}

x = \frac{- 2 \pm \sqrt{4 - 8 + 4 \sqrt{2})}}{2} \lor x = \frac{- 2 \pm \sqrt{4 - 8 - 4 \sqrt{2})}}{2}

x = \frac{- 2 \pm \sqrt{- 4 + 4 \sqrt{2})}}{2} \lor x = \frac{- 2 \pm \sqrt{- 4 - 4 \sqrt{2})}}{2}

x = \frac{- 2 \pm 2 \sqrt{\sqrt{2} - 1}}{2} \lor x = \frac{- 2 \mp 2 i \sqrt{\sqrt{2} + 1}}{2}

x = - 1 \pm \sqrt{\sqrt{2} - 1} \lor x = - 1 \mp i \sqrt{\sqrt{2} + 1}

y = \frac{- x}{x + 1}

y = \frac{- (- 1 \pm \sqrt{\sqrt{2} - 1})}{(- 1 \pm \sqrt{\sqrt{2} - 1}) + 1} \lor y = \frac{- (- 1 \mp i \sqrt{\sqrt{2} + 1})}{(- 1 \mp i \sqrt{\sqrt{2} + 1}) + 1}

y = \frac{1 \mp \sqrt{\sqrt{2} - 1}}{- 1 \pm \sqrt{\sqrt{2} - 1} + 1} \lor y = \frac{1 \pm i \sqrt{\sqrt{2} + 1}}{- 1 \mp i \sqrt{\sqrt{2} + 1} + 1}

y = \frac{1 \mp \sqrt{\sqrt{2} - 1}}{\pm \sqrt{\sqrt{2} - 1}} \times \frac{\pm \sqrt{\sqrt{2} - 1}}{\pm \sqrt{\sqrt{2} - 1}} \lor y = \frac{1 \pm i \sqrt{\sqrt{2} + 1}}{\mp i \sqrt{\sqrt{2} + 1}} \times \frac{\mp i \sqrt{\sqrt{2} + 1}}{\mp i \sqrt{\sqrt{2} + 1}}

y = \frac{(1 \mp \sqrt{\sqrt{2} - 1}) (\pm \sqrt{\sqrt{2} - 1})}{\sqrt{2} - 1} \lor y = \frac{(1 \pm i \sqrt{\sqrt{2} + 1}) (\mp i \sqrt{\sqrt{2} + 1})}{- (\sqrt{2} + 1)}

y = \frac{\pm \sqrt{\sqrt{2} - 1} \mp \sqrt{2} - 1)}{\sqrt{2} - 1} \lor y = \frac{\mp i \sqrt{\sqrt{2} + 1} - i^{2} \sqrt{2} + 1)}{- (\sqrt{2} + 1)}

y = \frac{\pm \sqrt{\sqrt{2} - 1} \mp \sqrt{2} - 1}{\sqrt{2} - 1} \lor y = \frac{\mp i \sqrt{\sqrt{2} + 1} + \sqrt{2} + 1}{- (\sqrt{2} + 1)}

Continue

Commented by tawa tawa last updated on 10/Jun/17

Am with you sir . God bless you sir .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

z^{2} + 2 (1 + j) z + {(1 + j)}^{2} = {(1 + j)}^{2} - 2 \Rightarrow

{(z + 1 + j)}^{2} = 1 + 2 j - 1 - 2 = 2 (j - 1)

2 (j - 1) = - 2 \sqrt{2} (\frac{\sqrt{2}}{2} - j \frac{\sqrt{2}}{2}) = - 2 \sqrt{2} e^{- j \frac{π}{4}}

\Rightarrow z + 1 + j = \pm j \sqrt[4]{8} e^{- j \frac{π}{8}} = \pm j \sqrt[4]{8} (c o s (- \frac{π}{8}) + j s i n (- \frac{π}{8})) =

= \pm j \sqrt[4]{8} (\frac{\sqrt{2 + \sqrt{2}}}{2} - j \frac{\sqrt{2 - \sqrt{2}}}{2}) =

= \pm 2^{\frac{3}{4}} \times 2^{- 1} \times 2^{\frac{1}{4}} (j \sqrt{\sqrt{2} + 1} + \sqrt{\sqrt{2} - 1}) =

= \pm (j \sqrt{\sqrt{2} + 1} + \sqrt{\sqrt{2} - 1})

\Rightarrow z = {\begin{cases} - 1 - j + j \sqrt{\sqrt{2} + 1} + \sqrt{\sqrt{2} - 1} \\ - 1 - j - j \sqrt{\sqrt{2} + 1} - \sqrt{\sqrt{2} - 1} \end{cases} =

\Rightarrow z = {\begin{cases} (\sqrt{\sqrt{2} - 1} - 1) + j (\sqrt{\sqrt{2} + 1} - 1) \\ - (\sqrt{\sqrt{2} - 1} + 1) - j (\sqrt{\sqrt{2} + 1} + 1) . \end{cases}

Commented by tawa tawa last updated on 10/Jun/17

God bless you sir .

Commented by tawa tawa last updated on 10/Jun/17

Please sir, am still trying to understand the workings sir .

please sir explain to me the steps . i mean the begining

Commented by tawa tawa last updated on 11/Jun/17

Sir i am lost from . \pm j (2^{\frac{3}{4}}) (\cos (- \frac{π}{8}) + jsin (- \frac{π}{8})]

why (- \frac{π}{8}) ? ? ? . How is it minus sir .

i thought it is \dots . \cos (\frac{π}{8}) - jsin (\frac{π}{4})

And how is it j (2^{3 / 4}) (\frac{\sqrt{2 + \sqrt{2}}}{2} - j \frac{\sqrt{2 - \sqrt{2}}}{2})

And the next steps \dots \dots

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

1) {(c o s x + j s i n x)}^{n} = c o s n x + j s i n n x

2) e^{j x} = c o s x + j s i n x

3) e^{- j x} = c o s (- x) + j s i n (- x) = c o s x - j s i n x

b e c a u s e : c o s (- x) = c o s x, s i n (- x) = - s i n x

\frac{\sqrt{2}}{2} - j \frac{\sqrt{2}}{2} = c o s (- \frac{π}{4}) + j s i n (- \frac{π}{4}) = e^{- j \frac{π}{4}}

a n g l e s i n c o s & s i n, s h o u l d b e t h e s a m e .

{(z + 1 + j)}^{2} = - 2 \sqrt{2} e^{- j \frac{π}{4}} = {(j)}^{2} \times 2^{\frac{3}{2}} \times e^{- j \frac{π}{4}}

\sqrt{{(z + 1 + j)}^{2}} = \pm \sqrt{{(j)}^{2} \times 2^{\frac{3}{2}} \times e^{- j \frac{π}{4}}}

z + 1 + j = \pm j \times 2^{\frac{3}{4}} \times e^{- j \frac{π}{8}}

e^{- j \frac{π}{8}} = c o s (- \frac{π}{8}) + j s i n (- \frac{π}{8}) = c o s (\frac{π}{8}) - j s i n (\frac{π}{8})

c o s \frac{π}{8} = \sqrt{\frac{1 + c o s \frac{π}{4}}{2} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}} = \frac{\sqrt{2 + \sqrt{2}}}{2} =

= \frac{\sqrt[4]{2} \sqrt{\sqrt{2} + 1}}{2} = 2^{\frac{1}{4}} \times 2^{- 1} \times \sqrt{\sqrt{2} + 1}

s i n \frac{π}{8} = \sqrt{\frac{1 - c o s \frac{π}{4}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2} =

= \frac{\sqrt[4]{2} \sqrt{\sqrt{2} - 1}}{2} = 2^{\frac{1}{4}} \times 2^{- 1} \times \sqrt{\sqrt{2} - 1}

\Rightarrow z + 1 + j = \pm j \times 2^{\frac{3}{4}} \times 2^{\frac{1}{4}} \times 2^{- 1} (\sqrt{\sqrt{2} + 1} - j \sqrt{\sqrt{2} - 1}) =

= \pm j \times 2^{\frac{3}{4} + \frac{1}{4} - 1} (\sqrt{\sqrt{2} + 1} - j \sqrt{\sqrt{2} - 1}) =

= \pm j \times 1 \times (\sqrt{\sqrt{2} + 1} - j \sqrt{\sqrt{2} - 1}) = \pm j (\dots . .)

i s i t o k ? m i s s t a w a ?

Commented by tawa tawa last updated on 11/Jun/17

Yes sir . i understand very well now . God bless you sir . i really appreciate

your time sir .