# Thread: More complex number questions

1. ## More complex number questions

Can help me to solve this?

4. If z=x+yi and $\displaystyle z^2=a+bi,x,y,a,b \in IR$, prove that
$\displaystyle 2x^2= \sqrt {(a^2+b^2)}+a$
By solving the equation $\displaystyle z^4+6z^2+25=0$ for $\displaystyle z^2$ or otherwise, express each root of the equation in the form x+yi.

5a. Express the complex numbers 1+i and 1-i in the form $\displaystyle r(cos \theta + i sin \theta )$, where r is the modulus and $\displaystyle \theta$ the argument of $\displaystyle \frac {(1+i)^5}{(1-i)^7}$.

5b. Given that $\displaystyle z_{1}=2+i$ and $\displaystyle z_{2}=-2+4i$, find, in the form a+bi, the complex number z such that $\displaystyle \frac {1}{z}= \frac {1}{z_{1}}+ \frac {1}{z_{2}}$.
Find the modulus and argument of z, giving your answer for arg z in the range $\displaystyle - \pi <argz \leq \pi$, correct to 3 decimal places.

4. $\displaystyle \pm(1+2i),\pm(1-2i)$
5a. $\displaystyle \sqrt{2}(cos\frac{\pi}{4}+i sin \frac{\pi}{4})$, $\displaystyle \sqrt{2}[cos(-\frac{\pi}{4})+i sin(-\frac{\pi}{4})]; \frac{1}{2},\pi$
5b. $\displaystyle \frac{6}{5}+\frac{8}{5}i; 2, 0.927 rad$

2. ## No.1

Ans for NO.1

$\displaystyle z=x+yi$, $\displaystyle z^2=x^2-y^2+2xyi....(1)$, $\displaystyle z^2=a+bi...(2)$
Hence, $\displaystyle a=x^2-y^2...(3$)
(from 2) $\displaystyle \bmod{z^2}=\sqrt{a^2+b^2}$,
(from 1)$\displaystyle \bmod{z^2}=\sqrt{(x^2-y^2)^2+)2xy)^2)}$$\displaystyle =\sqrt{x^4-2x^2y^2+y^4+4x^2y^2}$$\displaystyle =\sqrt{x^4+2x^2y^2+y^4}=x^2+y^2...(4)$

$\displaystyle a=x^2-y^2$
$\displaystyle \sqrt{a^2+b^2}=x^2+y^2$
$\displaystyle \sqrt{a^2+b^2}+a=2x^2$

Solve for $\displaystyle z^2$ by quadratic formula

$\displaystyle z^2=\frac{-6\pm\sqrt{36-100}}{2}=-3\pm4i$

Set $\displaystyle z= c+di$
$\displaystyle z^2=-3+4i$ or $\displaystyle z^2=-3-4i$

$\displaystyle z^2=c^2-d^2+2cdi$
Hence $\displaystyle c^2-d^2=-3$
$\displaystyle 2cd=\pm4$

$\displaystyle d=\pm\frac{2}{c}$

$\displaystyle c^2-(\pm\frac{2}{c})^2=-3$

$\displaystyle c^2-\frac{4}{c^2}=-3$

$\displaystyle c^4+3c^2-4=0$

$\displaystyle (c^2+4)(c^2-1)=0$

$\displaystyle c^2=-4, c^2=1$

$\displaystyle c=\pm2i,\pm1$

If $\displaystyle c=2i$, If $\displaystyle c=-2i$, If $\displaystyle c=1$, If $\displaystyle c=-1$
Then $\displaystyle d=\pm\frac{4}{4i}$, Then $\displaystyle d=\pm\frac{4}{-4i}$ ,, Then $\displaystyle d=\pm\frac{4}{2}$ ,, Then $\displaystyle d=\pm\frac{4}{-2}$
$\displaystyle d=\mp{i}$ ,, $\displaystyle d=\pm{i}$ ,, $\displaystyle d=\pm2$ ,, $\displaystyle d=\mp2$
$\displaystyle z=\pm(1+2i),\pm(1-2i)$

For number 5b, is it or $\displaystyle z_2=2+4i$?

3. Originally Posted by cloud5
5a. Express the complex numbers 1+i and 1-i in the form $\displaystyle r(cos \theta + i sin \theta )$, where r is the modulus and $\displaystyle \theta$ the argument of $\displaystyle \frac {(1+i)^5}{(1-i)^7}$.
Not sure what you're asking. I think it's easy to see that
$\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that
$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$.

DeMoivre's Theorem states that if
$\displaystyle z = r(\cos \theta + i \sin \theta )$

then
$\displaystyle z^n = r^n(\cos n\theta + i \sin n\theta )$.

So
\displaystyle \begin{aligned} (1 + i)^5 &= (\sqrt{2})^5 \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \\ &= 4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \end{aligned}

and
\displaystyle \begin{aligned} (1 - i)^7 &= (\sqrt{2})^7 \left(\cos \frac{49\pi}{4} + i\sin \frac{49\pi}{4}\right) \\ &= 8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right) \end{aligned}
(49pi/4 and pi/4 are coterminal angles.)

When you divide two complex numbers in polar form, you divide the two moduli and subtract the arguments:
\displaystyle \begin{aligned} \frac {(1+i)^5}{(1-i)^7} &= \frac{4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)}{8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)} \\ &= \frac{4\sqrt{2}}{8\sqrt{2}}\left[\cos \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) + i\sin \left(\frac{5\pi}{4} - \frac{\pi}{4}\right)\right] \\ &= \frac{1}{2}(\cos \pi + i \sin \pi) \\ &= -\frac{1}{2} \end{aligned}

01

4. Originally Posted by I-Think
For number 5b, is it or $\displaystyle z_2=2+4i$?
I can get the fraction but it contain the negative sign... Correction... $\displaystyle z_{2}=-2+4i$

5. Originally Posted by yeongil
Not sure what you're asking. I think it's easy to see that
$\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that
$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$.

DeMoivre's Theorem states that if
$\displaystyle z = r(\cos \theta + i \sin \theta )$

then
$\displaystyle z^n = r^n(\cos n\theta + i \sin n\theta )$.

So
\displaystyle \begin{aligned} (1 + i)^5 &= (\sqrt{2})^5 \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \\ &= 4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \end{aligned}

and
\displaystyle \begin{aligned} (1 - i)^7 &= (\sqrt{2})^7 \left(\cos \frac{49\pi}{4} + i\sin \frac{49\pi}{4}\right) \\ &= 8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right) \end{aligned}
(49pi/4 and pi/4 are coterminal angles.)

When you divide two complex numbers in polar form, you divide the two moduli and subtract the arguments:
\displaystyle \begin{aligned} \frac {(1+i)^5}{(1-i)^7} &= \frac{4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)}{8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)} \\ &= \frac{4\sqrt{2}}{8\sqrt{2}}\left[\cos \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) + i\sin \left(\frac{5\pi}{4} - \frac{\pi}{4}\right)\right] \\ &= \frac{1}{2}(\cos \pi + i \sin \pi) \\ &= -\frac{1}{2} \end{aligned}

01
Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this $\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that
$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$ explaination.

6. Originally Posted by cloud5
Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this $\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that
$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$ explaination.
$\displaystyle \frac{7\pi}{4}= \frac{2\pi}- \frac{-\pi}{4}$ so that $\displaystyle cos(\frac{7\pi}{4}= cos(\frac{-\pi}{4})= cos(\frac{\pi}{4})$ and $\displaystyle sin(\frac{7\pi}{4})= sin(-\frac{\pi}{4})= -sin(\frac{\pi}{4})$.
Was that what was bothering you?

7. Originally Posted by cloud5
I can get the fraction but it contain the negative sign... Correction... $\displaystyle z_{2}=-2+4i$
I think the answer given is not correct .

$\displaystyle \frac{1}{z} =\frac{1}{z_1}+\frac{1}{z_2}$

$\displaystyle =\frac{1}{2+i}+\frac{1}{-2+4i}$

$\displaystyle =\frac{5i}{-8+6i}$

$\displaystyle z=\frac{6}{5}-\frac{8}{5}i$

8. Originally Posted by cloud5
Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this $\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that
$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$ explaination.
For a complex number $\displaystyle a + bi$, to convert to polar form $\displaystyle r(\cos \theta + \sin \theta)$, you need to know the following:
$\displaystyle r = \sqrt{a^2 + b^2}$
$\displaystyle \theta = \tan^{-1}\left(\frac{b}{a}\right)$ (subject to correction, since inverse tangent only gives angles in $\displaystyle -\pi/2 < \theta < \pi/2$.

For 1 + i (or 1 + 1i),
$\displaystyle r = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\displaystyle \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$

For 1 - i (or 1 - 1i),
$\displaystyle r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
$\displaystyle \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}$
As HallsofIvy has shown, $\displaystyle -\frac{\pi}{4}$ & $\displaystyle \frac{7\pi}{4}$ are coterminal angles. I added $\displaystyle 2\pi$ above because I wanted an angle that was in the interval $\displaystyle [0, 2\pi)$.

As far as I'm concerned my answers match what you originally posted.
5a. $\displaystyle \sqrt{2}(cos\frac{\pi}{4}+i sin \frac{\pi}{4})$, $\displaystyle \sqrt{2}[cos(-\frac{\pi}{4})+i sin(-\frac{\pi}{4})]; \frac{1}{2},\pi$
The last two numbers, $\displaystyle \frac{1}{2} \text{and} \pi$, are the modulus and argument, respectively, of my answer
$\displaystyle \frac{1}{2}(\cos \pi + i \sin \pi)$.

01

9. There is a very simple way to work 5a.
This is a grossly underused identity. $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$

Notice that $\displaystyle \left( {1 + i} \right)^5 \left( {1 + i} \right)^7 = \left( {1 + i} \right)^{12} = \left( {{\sqrt 2 }\exp (\pi i/4)} \right)^{12} = - 2^6$

So $\displaystyle \frac{{\left( {1 + i} \right)^5 }}{{\left( {1 - i} \right)^7 }} = \frac{{\left( {1 + i} \right)^{12} }}{{\left( {\sqrt 2 } \right)^{14} }} = \frac{{ - 2^6 }}{{2^7 }} = \frac{{ - 1}}{2}$.

10. Originally Posted by HallsofIvy
$\displaystyle \frac{7\pi}{4}= \frac{2\pi}- \frac{-\pi}{4}$ so that $\displaystyle cos(\frac{7\pi}{4}= cos(\frac{-\pi}{4})= cos(\frac{\pi}{4})$ and $\displaystyle sin(\frac{7\pi}{4})= sin(-\frac{\pi}{4})= -sin(\frac{\pi}{4})$.
Was that what was bothering you?
After looking at yeongil's explanation, I finally got the point. Thanks for the help~

$\displaystyle z=\frac{6}{5}-\frac{8}{5}i$
I got that too, but how to change $\displaystyle \frac{1}{z}$to z?

12. Originally Posted by cloud5
I got that too, but how to change $\displaystyle \frac{1}{z}$to z?
You mean from this:
$\displaystyle \frac{1}{z}=\frac{5i}{-8+6i}$
to this?
$\displaystyle z=\frac{6}{5}-\frac{8}{5}i$

Just take the reciprocal of both sides and multiply top and bottom by -5i:
\displaystyle \begin{aligned} \frac{1}{z} &= \frac{5i}{-8+6i} \\ \frac{z}{1} &= \frac{-8+6i}{5i} \\ \end{aligned}
\displaystyle \begin{aligned} z &= \frac{-8+6i}{5i} \times \frac{-5i}{-5i} \\ &= \frac{40i + 30}{25} \\ &= \frac{30}{25} + \frac{40}{25}i \\ &= \frac{6}{5} + \frac{8}{5}i \end{aligned}

Umm, mathaddict, I'm not getting what you got. In fact, this matches cloud5's answer in the original post.

01

13. Originally Posted by yeongil
You mean from this:
$\displaystyle \frac{1}{z}=\frac{5i}{-8+6i}$
to this?
$\displaystyle z=\frac{6}{5}-\frac{8}{5}i$

Just take the reciprocal of both sides and multiply top and bottom by -5i:
\displaystyle \begin{aligned} \frac{1}{z} &= \frac{5i}{-8+6i} \\ \frac{z}{1} &= \frac{-8+6i}{5i} \\ \end{aligned}
\displaystyle \begin{aligned} z &= \frac{-8+6i}{5i} \times \frac{-5i}{-5i} \\ &= \frac{40i + 30}{25} \\ &= \frac{30}{25} + \frac{40}{25}i \\ &= \frac{6}{5} + \frac{8}{5}i \end{aligned}

01
Yeah! Thank you!