find the complex number that satisfies the following conditions

Question

Find all values of $z \in \Bbb C$ such that: $z + \bar{z} = 18$ and $z.\bar{z} = 84$.

I don't know how to get that values, someone can help me to solve this?

Do you mean $z + \bar{z} = 18$ and $z \cdot \bar{z} = 84$? – Calvin Lin Jun 07 '13 at 00:53 — Calvin Lin, Jun 07 '13 at 00:53

score 2 · Answer 1 · answered Jun 07 '13 at 00:51

2

The only complex number that satisfies $z + z = 2z = 18$ is $z = 9$, but $9^2 = 81$, so there are no values that satisfy both conditions.

answered Jun 07 '13 at 00:51

user81310

21

1

Look again the question, please – Tomi Jun 07 '13 at 01:02

score 2 · Answer 2 · edited Jun 07 '13 at 01:39

2

If $z = a+bi$ then the first equation results in $2a = 18$ so $a = 9$. The second equation results in $a² + b² = 84$ with $a = 9$ gives $b = \pm \sqrt{3}$. So there are 2 solutions: $9 + \sqrt{3}i$ and its conjugate $9 - \sqrt{3}i$. No heavy algebra needed!

edited Jun 07 '13 at 01:39

Oleg567

17,295

answered Jun 07 '13 at 01:18

imranfat

10,029

thanks Oleg, I realized the i was missing – imranfat Jun 07 '13 at 01:40

Tapu · Accepted Answer · 2013-06-07T01:08:10.237

1

Hint: (Please check first that you have written the question correctly, because $z+z$ looks odd)

Any complex number can be written as $z=a+i b$ where $a,b$ are real numbers. Substitute into the given conditions,...., and solve for $a,b$.

Edit:

Now your question makes sense. Try this: $$z.\bar{z} =84$$ $$\Rightarrow z.(18-z) =84$$ $$z^2-18z+84=0$$ This is a quadratic equation. So....

edited Jun 07 '13 at 01:08

answered Jun 07 '13 at 00:49

Tapu

3,496

It's $z+z$, yes – Tomi Jun 07 '13 at 00:53
1

And if the second one is also $z.z=84$, then please accept @user81310's answer. – Tapu Jun 07 '13 at 00:55

score 0 · Answer 4 · answered Jun 07 '13 at 01:02

$z$ and $\bar{z}$ are roots to the quadratic equation $X^2 - 18X + 84 = 0 $. By the quadratic formula, the roots are

$$ X = \frac{ 18 \pm \sqrt{ 18^2 - 4 \times 84 } } { 2 } = 9 \pm \sqrt{3} i$$

Note, you could do it slowly by setting $ z= a+bi$ and then solving for real values of $a$ and $b$. This gives you a direct immediate approach.

score 0 · Answer 5 · answered Jun 07 '13 at 01:44

$z + \bar{z} = 18$ and $z \bar{z} = 84$

.$z + \bar{z} = 2Re(z)$

so(...)

$z + \bar{z} = 18 \rightarrow 2a = 18 \rightarrow a = 9$

. $z \bar{z} = |z|^2$

so(...)

$\left(\sqrt{9^2+b^2}\right)^2 = 84 \rightarrow 81 + b^2 = 84 \rightarrow b^2 = 3 \rightarrow b = \pm\sqrt{3}$

So, the solutions:

$9+\sqrt3i$ & $9-\sqrt3i$

find the complex number that satisfies the following conditions

5 Answers5