# Thread: More complex number questions

1. ## More complex number questions

Can help me to solve this?

4. If z=x+yi and $z^2=a+bi,x,y,a,b \in IR$, prove that
$2x^2= \sqrt {(a^2+b^2)}+a$
By solving the equation $z^4+6z^2+25=0$ for $z^2$ 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 $r(cos \theta + i sin \theta )$, where r is the modulus and $\theta$ the argument of $\frac {(1+i)^5}{(1-i)^7}$.

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

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

2. ## No.1

Ans for NO.1

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

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

Solve for $z^2$ by quadratic formula

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

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

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

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

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

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

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

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

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

$c=\pm2i,\pm1$

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

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

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

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

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

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

So
\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
\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:
\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 $z_2=2+4i$?
I can get the fraction but it contain the negative sign... Correction... $z_{2}=-2+4i$

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

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

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

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

So
\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
\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:
\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 $1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

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

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

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

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

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

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

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

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 $1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

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

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

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

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

01

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

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

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

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

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

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

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

\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:
$\frac{1}{z}=\frac{5i}{-8+6i}$
to this?
$z=\frac{6}{5}-\frac{8}{5}i$

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

\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!