# Complex number

• Feb 14th 2011, 05:52 AM
MichaelLight
Complex number
Under the section of complex number, i faced 2 questions which i couldn't answer... Here they go...

-Show that x=1-2i is a root of the equation x3-3x2+7x-5=0. Hence, find all the roots of the equation.

-Express Attachment 20803 in the form a+ib, where a>0

Note: for both question, 'i' represents imaginary number... Please show step-by-step working if possible... Thanks...
• Feb 14th 2011, 05:56 AM
Prove It
Q1 First note that $\displaystyle x^3 - 3x^2 + 7x - 5 = 0$ when $\displaystyle x=1$, so $\displaystyle (x-1)$ is a factor...

Long dividing gives the quadratic factor $\displaystyle x^3 - 3x^2 + 7x - 5 = (x - 1)(x^2 - 2x + 5)$.

Now factorise the quadratic factor to get the required result.

Q2: Express $\displaystyle 1 + i\sqrt{3}$ in polar form, then use DeMoivre's Theorem.
• Feb 14th 2011, 06:37 AM
TheCoffeeMachine
Quote:

Originally Posted by MichaelLight
-Show that x=1-2i is a root of the equation $x^3-3x^2+7x-5 = 0$. Hence, find all the roots of the equation.

• Show that $x^3-3x^2+7x-5 = 0$ yields zero at $x = 1-2i$.
• Hint for finding the rest of the roots: note that $1+2i$ is also a root.
• For the second question, use De Moivre's theorem, as Prove It suggested.

Quote:

Please show step-by-step working if possible... Thanks..
Notes like this usually serve as a reminder for me not to spoon-feed detailed solutions! (Giggle)
• Feb 14th 2011, 06:45 AM
Quote:

Originally Posted by MichaelLight
Under the section of complex number, i faced 2 questions which i couldn't answer... Here they go...

-Show that x=1-2i is a root of the equation x3-3x2+7x-5=0. Hence, find all the roots of the equation.

-Express Attachment 20803 in the form a+ib, where a>0

Note: for both question, 'i' represents imaginary number... Please show step-by-step working if possible... Thanks...

Since the polynomial coefficients are real...

$x=1-2i$ is a root

$\Rightarrow\ x=1+2i$ is also a root

$(x-1+2i)(x-1-2i)(x-\alpha)=x^3-3x^2+7x-5$

$\left(x^2-x-2ix-x+1+2i+2ix-2i-4i^2\right)(x-\alpha)=x^3-3x^2+7x-5$

$\left(x^2-2x+5\right)(x-\alpha)=x^3-3x^2+7x-5$

which gives $\alpha=1$ from the product of the last 2 terms

$(a+ib)^2=1+i\sqrt{3}\Rightarrow\ a^2-b^2=1$

and $2ab=\sqrt{3}$

Hence solve the simultaneous equations

$a=\frac{\sqrt{3}}{2b}$

$\frac{3}{4b^2}-b^2=1\Rightarrow\ 3-4b^4=4b^2\Rightarrow\ 4b^4+4b^2-3=0$

$\left(2b^2-1\right)\left(2b^2+3\right)=0$

and since "b" is real

$2b^2=1\Rightarrow\ b^2=\frac{1}{2}\Rightarrow\ b=\pm\frac{1}{\sqrt{2}}$

$\Rightarrow\ a=\pm\sqrt{\frac{3}{2}}$

while "a" and "b" must have same signs as $2ab=\sqrt{3}$
• Feb 14th 2011, 07:33 AM
FernandoRevilla
Quote:

Originally Posted by MichaelLight
-Express $\sqrt{1+i\sqrt{3}}$ in the form a+ib, where a>0

Another way (without using polar form):

$\sqrt{1+i\sqrt{3}}=a+bi\Leftrightarrow (a+bi)^2=1+i\sqrt{3}\Leftrightarrow a^2-b^2+2abi=1+i\sqrt{3}\Leftrightarrow$

$\begin{Bmatrix} a^2-b^2=1\\2ab=\sqrt{3}\end{matrix}\Leftrightarrow\;\l dots$

Fernando Revilla

Edited: Sorry, now I see that the problem was already solved this way.