0
$\begingroup$

How to determine real and imaginary part of $z^{2} -7z + (13+i)=0$ ?

Should i begin with placing $a+bi$ instead of $z$, and if so, how should i continue calculating?

$\endgroup$
  • $\begingroup$ You have given an equation but you are asking find the real and imaginary parts of it. Do you mean how to find the real and imaginary parts of the left hand side? So you can then compare it to the right hand side? $\endgroup$ – Ian Miller Nov 6 '16 at 0:53
  • $\begingroup$ I believe that it should be solved as any regular quadratic equation, just at the end we should get complex number as a result and we need to determine real and imaginary part. $\endgroup$ – Tars Nolan Nov 6 '16 at 0:56
  • $\begingroup$ @lala The first step to answering a problem is to work out what you are actually meant to do. $\endgroup$ – Ian Miller Nov 6 '16 at 0:57
  • 1
    $\begingroup$ Your question is unclear. Usually, an equation of the form $\cdots=0$ is associated with a "solve the following equation". There is a very simple silly answer to your question if I read it "determine real and imaginary part of $0$ : answer 0 and 0 ! So, please, make your text not ambiguous. $\endgroup$ – Jean Marie Nov 6 '16 at 1:09
1
$\begingroup$

The quadratic formula applies for complex numbers also! So just apply it: $$ z = \frac{7 \pm \sqrt{49 -52-4i}}{2} = \frac{7 \pm \sqrt{-3-4i}}{2} $$

We can calculate the square root of $-3-4i$, it is $1 - 2i$ (I shall edit the question if you want to know how). So the answer simplifies to: $$ z = \frac{7 \pm (1 - 2i)}{2} = 4-i/3+i $$

Hence, these are the two values of $z$. The real parts of the roots are $4,3$ respectively, and the imaginary parts are $-1,1$ respectively.

EDIT: There are two ways of deriving the square root of $-3-4i$. One is to imagine that $(a+bi)^2 = -3-4i$. This expands to $a^2-b^2 + 2abi = -3-4i$. Comparing real parts, we get two equations: $$ a^2-b^2 = 3; 2ab = -4 $$

Here, we use a small trick: the following identity (called the Pythagorean triplet identity) $$ (a^2-b^2)^2 + (2ab)^2 = (a^2+b^2)^2 $$

From this identity, we see that $(a^2+b^2) = \sqrt{3^2 + (-4)^2} = 5$ (the negative square root cannot come as $a,b$ are real ,so $a^2+b^2$ is non-negative).

So, we know $a^2+b^2=5$ and $a^2-b^2=3$. Hence, adding them, $2a^2 = 8$ and $a^2=4$. Subtracting the second from the first, $2b^2=2$ so $b^2=1$. Now, observe that $ab$ is negative, hence $a$ and $b$ must have opposing signs. That leads to two possibilities, which are additive inverses of each other: $a=1;b=-2$ and $a=-1,b=2$, leading to $1-2i$ and $-1+2i$ as the square roots.

There is another method, where you can rewrite the given complex number in polar form and use De Moivre's theorem to derive any power of the complex number, including the square root.

$\endgroup$
  • $\begingroup$ Feel free to edit, it would be very useful to see how can i get square root from $-3-4i$. $\endgroup$ – Tars Nolan Nov 6 '16 at 1:01
  • $\begingroup$ @JeanMarie You are right. To be fair, when I posted the answer I expected that the roots, if not entirely answer the question, could at least help in answering it. That's why I posted at least that much, so it is a forward step if not an answer. But what else can you do with such an equation? $\endgroup$ – астон вілла олоф мэллбэрг Nov 6 '16 at 1:12
  • 1
    $\begingroup$ @TarsNolan Question edited. $\endgroup$ – астон вілла олоф мэллбэрг Nov 6 '16 at 1:13
1
$\begingroup$

$$z^2 -7z + (13+i)=0$$ $$\text{Let }z=a+bi\text{ where }a,b\in\mathbb{R}$$ $$(a+bi)^2-7(a+bi)+13+i=0$$ $$a^2+2abi-b^2-7a-7bi+13+i=0$$ As $a,b\in\mathbb{R}$ you can now compare the real and imaginary parts of both sides to solve the equation: $$a^2-b^2-7a+13=0\text{ and }2ab-7b+1=0$$ $$\text{The second equation gives: }b=\frac{1}{7-2a}$$ $$a^2-\left(\frac{1}{7-2a}\right)^2-7a+13=0$$ $$a^2(7-2a)^2-1-7a(7-2a)^2+13(7-2a)^2=0$$ $$4a^4-56a^3+297a^3-707a+636=0$$ $$(a-3)(a-4)(4a^2-28a+53)=0$$

$$\text{So }a=3\text{ or }a=4$$

Substituting this back into $b=\frac{1}{7-2a}$ gives $b=\pm1$. So the answers are:

$$z=3+i\text{ and }z=4-i$$

Note: that this is not the easiest way to solve the quadratic equation you gave but you asked to evaluate real and imaginary parts.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.