# Math Help - [SOLVED] easy complex number questions

1. ## [SOLVED] easy complex number questions

hi, could someone check the answers of these for me? btw, the book says that they are all wrong. i dont understand why.

1. If z = 2 + i, evaluate:

a) 3z = 6 + 3i
b) $(z-1)(z^2 + z +1)$
$= z^3 - 1$
$= (2+i)^3$
$= (2 + i)^2(2+i)$
$= (4 + i^2 + 4i)(2+2i)$
$= (3 + 4i) (2+2i)$
$= 6 + 8i + 6i + 8i^2$
$= -2 + 14i$

2. Solve for z; expressing answers in the form a + ib
a) $\frac{z+3}{z - 1} = 2 - 3i$
$\frac{z -1 + 4}{z-1} = 2-3i$
$1 + \frac{4}{z-1} = 2 - 3i$
$2 - 3i -1 = \frac{4}{z-1}$
$1 - 3i = \frac{4}{z -1}$
$\frac{4}{1-3i} + 1 = z$
$\frac{4(1 + 3i)}{(1-3i)(1+3i)}+ 1 =z$
$\frac{4 + 12i}{10} + 1 = z$
$\frac{4+12i + 10}{10} = z$
$\frac{14 + 12i}{10} = z$
$\frac{7}{5} + \frac{6}{5}i = z$

Solve for x. $\frac{2z}{2+i} + 3- 2i = (1-i)z$

3. Simplify: $\frac{\sqrt{5+12i}+ \sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}$

I got 3i/2. but the answer is -3i/2. I dont get why.

Thanks

2. ## [SOLVED] solving complex number equation

Solve for z. $\frac{2z}{2+i} + 3- 2i = (1-i)z$

I tried it but nothing gets rid of the z.. could someone give me a hint? I multiplied by 2 + i on both sides, but that didn't work. did a whole lot of algebra bashing, but nothing gets rid of the 2z on the left hand side.

HI

Let $z = a + ib$

$\frac{2(a+bi)}{2+i}+3-2i=(1-i)(a+bi)$

Multiply by its conjugate,

$
\frac{(2a+2bi)(2-i)}{5}+3-2i=(a+b)+(b-a)i
$

$4a+2b+15-2ai+4bi-10i=5(a+b)+5(b-a)i$

By comparing ,

$4a+2b+15=5(a+b)$.... 1

$4b-2a-10=5(b-a)$.... 2

Now solve the system for a and b
..

3. Originally Posted by differentiate

Solve for x. $\frac{2z}{2+i} + 3- 2i = (1-i)z$

3. Simplify: $\frac{\sqrt{5+12i}+ \sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}$

I got 3i/2. but the answer is -3i/2. I dont get why.

Thanks
your sum 1(a), 1(b),2(a) are right
but 2(b) solve for x
$\frac{2z}{2+i} + 3- 2i = (1-i)z$
where is x (?) in the statement ?

4. Help me find the mistake in my working out. I got 3i/2, however the answer is -3i/2.

Evaluate: $\frac{\sqrt{5+12i} + {\sqrt{5-2i}}}{\sqrt{5+12i} - {\sqrt{5-12i}}}$

$\sqrt{5+12i} = a + bi$
$5 + 12i = a^2 + b^2i^2 + 2abi$
$5 + 12i = a^2 - b^2 + 2abi$
$2ab = 12, a^2 - b^2 = 5$
$b = 6/a$
$a^2 - \frac{6^2}{a^2} = 5$
$a^4 - 5a^2 - 36 = 0$
$(a^2 - 9)(a^2 +4) = 0$
$a^2 = \pm3$
$\sqrt{5+12i} = \pm (3+12i)$

$\sqrt{5 - 12i} = a + bi$
$5 - 12i = a^2 + b^2i^2 + 2abi$
$5 - 12i = a^2 - b^2 + 2abi$
$2ab = -12, a^2 - b^2 = 5$
$b = \frac{-6}{a}$
$a^2 - (\frac{-6}{a})^2 = 5$
$a^4 - 5a^2 - 36 = 0$
$(a^2 - 9)(a^2 + 4)=0$
$a = \pm 3$
$\sqrt{5 - 12i} = \pm (3-2i)$

hence, $\frac{\sqrt{5+12i} + {\sqrt{5-2i}}}{\sqrt{5+12i} - {\sqrt{5-12i}}}$
$= \frac{3 + 2i + 3 - 2i}{3 + 2i - 3 + 2i} = \frac{6}{4i} = \frac{3}{2i}$

5. You have

$\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}} = \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}\cdot \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}+\sqrt{5-12i}}$

This is just

$\frac{5+12i + 2 \sqrt{5-12i}\sqrt{5+12i}+5-12i}{5+12i-5+12i}$

Try this -- you might find it easier.

6. Originally Posted by harbottle
You have

$\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}} = \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}\cdot \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}+\sqrt{5-12i}}$

This is just

$\frac{5+12i + 2 \sqrt{5-12i}\sqrt{5+12i}+5-12i}{5+12i-5+12i}$

Try this -- you might find it easier.
Then to perform the complex division so you don't have a complex denominator, multiply top and bottom by the bottom's conjugate.

7. alright so then $\frac{10 + 2\sqrt{169}}{24i}$
$\frac{36}{24i}$
$\frac{36i}{24i^2}$
$\frac{-3}{2}$

8. Originally Posted by differentiate
alright so then $\frac{10 + 2\sqrt{169}}{24i}$
$\frac{36}{24i}$
$\frac{36i}{24i^2}$
$\frac{-3}{2}$

okay, so my question is... the method I used before.. like by finding the square roots and stuff. Why didn't that work?
Your working out is correct here...