# complex number help needed

• Jun 20th 2006, 03:54 AM
watsonmath
complex number help needed
Need help in solving these complex numbers i tried hard but cant figure out the working. And anyone good in complex number give me tip&guideline in mastering this topic???

2. if Z = 5-i2, find the real and imaginary parts of:
d) (Z-1)/(Z-i)

3. Find real numbers x and y such that:
c) (1+i)x-(2-i3)y = 10 Ans: x = 6 ; y = -2
d) (2x-i3y) - (1 + i5)x = 3 + i2 Ans:x = 3 ; y = -17/3

7. Simplify the following and express the answer in polar form:
c) (2-i3)/(3+i2) - (3+i)/(4+i3) Ans:0.37<-28.8 degree
e) (2-i3)(3-i2)/i(1+i) Ans:9.17<225 degree

Thanks alot.......
• Jun 20th 2006, 08:41 AM
Soroban
Hello, watsonmath!

Here are a few of them . . .

Quote:

2d) If $Z = 5 - 2i$, find the real and imaginary parts of: . $\frac{Z-1}{Z-i}$
We have: . $\frac{Z - 1}{Z - i}\;=\;\frac{(5 - 2i) - 1}{(5 - 2i) - i}\;=\;\frac{4 - 2i}{5 - 3i}$

Rationalize: . $\frac{4 - 2i}{5 - 3i}\cdot\frac{5 + 3i}{5 + 3i}\;=\;\frac{20 + 12i - 10i - 6i^2}{25 + 15i - 15i - 9i^2}\;=\;\frac{20 + 12i - 10i + 6}{25 + 9}$

Therefore: . $\frac{26 + 2i}{34} \;= \;\frac{26}{34} + \frac{2i}{34}\;=\;\frac{13}{17} + \frac{1}{17}i$

Quote:

3. Find real numbers x and y such that:
$c)\;(1 + i)x - (2 -3i)y \:= \:10$ . . . Ans: $x = 6,\;y = -2$
We have: . $x + xi - 2y - 3yi \;=\;10$

Then: . $(x - 2y) + (x + 3y)i \;=\;10$

Two complex numbers are equal if their real components are equal
. . and their imaginary components are equal.

So we have: . $\left\{\begin{array}{cc} x - 2y\:=\:10 \\ x+ 3y\:=\:0\end{array}$

Solve the system and get: . $x = 6,\;y = -2$

Quote:

$d)\;(2x - 3yi) - (1 + 5i)x \:= \:3 + 2i$ . . . Ans: $x = 3,\;y = -\frac{17}{3}$
We have: . $2x - 3yi - x - 5xi \;= \;3 + 2i$

Then: . $x + (-5x - 3y)i\;=\;3 + 2i$

Equate real and imaginary components: . $\left\{\begin{array}{cc}x \:= \:3 \\ -5x - 3y \:=\:2\end{array}$

And we get: . $x = 3,\;y = -\frac{17}{3}$
• Jun 21st 2006, 06:03 AM
watsonmath
thanks Soroban!

can you explain the last 2 for me too pls....... i tired hard but my answer were wrong or close to.

foe eg. qns 7c, my r=0.37 but my degree is different from the answer given in my book. so i need your help to check if i am wrong or the book made a mistake.
• Jun 21st 2006, 06:23 AM
Quick
Quote:

Originally Posted by Soroban
Rationalize: . $\frac{4 - 2i}{5 - 3i}\cdot\frac{5 + 3i}{5 + 3i}\;=\;\frac{20 + 12i - 10i - 6i^2}{25 + 15i - 15i - 9i^2}\;=\;\frac{20 + 12i - 10i + 6}{25 + 9}$

I don't think this is right becuase
$\frac{20 + 12i - 10i - 6i^2}{25 + 15i - 15i - 9i^2}$ $=\frac{20 + 12i - 10i - 6i^2}{25 + \not{15i} - \not{15i} - 9\not{i^2}}$ $=\frac{20 + 12i - 10i - 6i^2}{25 + (-9)}$ $=\frac{20 + 12i - 10i - 6i^2}{25 - 9} \neq\frac{20 + 12i - 10i - 6i^2}{25 + 9}$
or am I missing something?
• Jun 21st 2006, 06:38 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
I don't think this is right becuase
$\frac{20 + 12i - 10i - 6i^2}{25 + 15i - 15i - 9i^2}$ $=\frac{20 + 12i - 10i - 6i^2}{25 + \not{15i} - \not{15i} - 9\not{i^2}}$ $=\frac{20 + 12i - 10i - 6i^2}{25 + (-9)}$ $=\frac{20 + 12i - 10i - 6i^2}{25 - 9} \neq\frac{20 + 12i - 10i - 6i^2}{25 + 9}$
or am I missing something?

$i^2=-1$
• Jun 21st 2006, 07:41 AM
Soroban
Hello, watsonmath!

I have issues with the answers to #7 . . .

Quote:

7. Simplify the following and express the answer in polar form:

$c)\;\;\frac{2 -3i}{3+2i} - \frac{3+i}{4+3i}$ . . . Ans: $0.37,\;-28.8^o$ ??

$e)\;\frac{(2-3i)(3-2i)}{i(1+i)}$ . . . Ans: $9.17,\;225^o$ ??

$7c)$ Rationalize: . $z \;= \;\frac{2-3i}{3+2i}\cdot\frac{3-2i}{3-2i} - \frac{3+i}{4+3i}\cdot\frac{4-3i}{4-3i}$

. . $z \;= \;\frac{6 - 4i - 9i + 6i^2}{9 - 4i^2} - \frac{12 - 9i + 4i - 3i^2}{16 - 9i^2} \;=\;\frac{-13i}{13} - \frac{15 - 5i}{25}$

. . $z \;= \;-i - \frac{3 - i}{5}\;=\;-\frac{3}{5} - \frac{4}{5}i$

Then: . $r\;=\;|z|\;=\;\sqrt{\left(-\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} \;= \;\sqrt{\frac{9}{25} + \frac{1}{25}} \;= \;\sqrt{\frac{25}{25}}\;=\;1$

And: . $\tan\theta\;=\;\frac{-\frac{4}{5}}{-\frac{3}{5}}\;=\;\frac{4}{3}\quad\Rightarrow\quad \theta \:=\:53.1^o,\;233.1^o$

Since $z$ is in quadrant 3: . $\theta = 233.1^o$

Polar form: . $z\:=\:\cos233.1^o + i\sin233.1^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$7e)$ Multiply: . $z \;= \;\frac{(2-3i)(3-2i)}{i(1+i)}\;=\;\frac{6 - 4i - 9i + 6i^2}{-1 + i} \;= \;\frac{-13i}{-1 + i}$

Rationalize: . $z \;= \;\frac{-13i}{-1 + i}\cdot\frac{-1 -i}{-1-i}\;=\;\frac{13i + 13i^2}{1 - i^2} \;= \;$ $\frac{-13 + 13i}{2} \;= \;-\frac{13}{2} + \frac{13}{2}i$

Then: . $r\;=\;|z|\;=\;\sqrt{\left(-\frac{13}{2}\right)^2 + \left(\frac{13}{2}\right)^2}\;=\;$ $\sqrt{2\left( \frac{13}{2}\right)^2}\;=\;\frac{13}{2}\sqrt{2} \; \approx \; 9.19$

And: . $\tan\theta \;= \;\frac{\frac{13}{2}}{\text{-}\frac{13}{2}}\;=\;-1\quad\Rightarrow\quad\theta\:=\:-45^o,\;135^o$

Since $z$ is in quadrant 2: . $\theta = 135^o$

Polar form: . $z\;=\;9.19\left(\cos135^o + i\sin135^o\right)$

• Jun 21st 2006, 07:53 AM
watsonmath
thanks alot Soroban..

so my book has the wrong answer to it???