# modulus-argument form

• October 1st 2008, 08:16 PM
Dr Zoidburg
modulus-argument form
Let $z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$

(a) express z in modulus-argument form

(b) express z in real-imaginary form

First Question: Do I need to expand out the top and bottom parts before doing anything else?

Edit: This is what I got for (b):
expanding the denominator out, we get (-2 + 2i)
multiplying by $\frac{(-2 - 2i)}{(-2 - 2i)}$ gives $\frac{(4i - 4)(1 + \sqrt{3i})^3}{8}$ = $\frac{(i-1)(1 + \sqrt{3i})^3}{2}$
expanding out $(1 + \sqrt{3i})^3$ gives:
$[1 + 2\sqrt{3i} + 3i](1 + \sqrt{3i}) = 1 + 3\sqrt{3i} + 9i + 3i\sqrt{3i}$
$3i\sqrt{3i} = 3\sqrt{3i^3} = -3\sqrt{3i}$ thus:
$= 1 + 3\sqrt{3i} + 9i - 3\sqrt{3i} = 1 + 9i$
hence:
$z = \frac{(i -1)(1 + 9i)}{2}$
$z = \frac{(-10 -8i)}{2}$
$z = -5 - 4i$

correct? yes/no?

Any help for part (a) greatly appreciated!

FURTHER EDIT:
I've been through Moo's workings, adding in the cube power he missed and got the following:
(using Moo's notation of x, y, & t)
$x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{2}=2 e^{-i \tfrac \pi {2}}}$
${y=2 e^{i \tfrac \pi 3}}\implies \boxed{y^{3}=8 e^{i \pi} = -8}$
$t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{3}=2 \sqrt{2} e^{i \tfrac{{3}\pi} 4}}$

which gives:
$
z=\frac{2 e^{-i \tfrac \pi 2} \cdot -8}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}} \implies
-4\sqrt{2} \cdot \frac{e^{-i \tfrac \pi 2}}{e^{i \tfrac{3\pi} 4}} \implies -4\sqrt{2} \cdot e^{i (\tfrac {-\pi} {2} - \tfrac {3\pi} {4})} \implies 4\sqrt{2} \cdot e^{-i \tfrac {5\pi} {4}}
$

$
z= -4\sqrt{2} (cos \tfrac {-5\pi} {4} + isin \tfrac {-5\pi} {4}) \implies-4\sqrt{2}(-\tfrac {1} {\sqrt{2}} + i \tfrac {1} {\sqrt{2}})$

$
z= 4 - 4i$

which is ever-so-slightly different to what I got by expanding it all out.

What the hell am I doing wrong here?! I've spent 1/2 the morning going over this and just can't for the life of me see where I've gone wrong. Any help desperately appreciated!
• October 1st 2008, 08:26 PM
icemanfan
I'm not sure what modulus-argument form is, but in order to express this number as $a + bi$, I would multiply by $\frac{(1 - i)^3}{(1 - i)^3}$ in order to get a real denominator. Then you can multiply out the top part and simplify it and everything will work out. It's a bit tedious, but I can't think of an easier way of doing it.
• October 1st 2008, 08:49 PM
Dr Zoidburg
That's what I figured for the 2nd part as well. Though it's a little easier to expand the bottom out first, as that gives $(-2 + 2i)$ and then muliply both by $(-2 - 2i)$ which gives us 8 as the demoninator.

I hope.
• October 1st 2008, 11:08 PM
Moo
Hello,

Yes, it gives 8. But you don't need to do it this way, you can keep all the original stuff.

Remember that $e^{ix}=\cos(x)+\sin(x) i$

$z=\frac{(1-i)^2 (1+\sqrt{3} i)}{(1+i)^3}=\frac{x^2 \cdot y}{t^3}$
In order to have the modulus argument form, you have to divide each complex part by its modulus.

$x=1-i \implies |x|=\sqrt{1+1}=\sqrt{2}$
$\frac{x}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i=e^{-i \tfrac \pi 4}$

$x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{\color{red}2}=2 e^{-i \tfrac \pi {\color{red}2}}}$

$y=1+\sqrt{3} i \implies |y|=\sqrt{1+(\sqrt{3})^2}=2$
So we have $\frac{y}{2}=\frac{1+\sqrt{3} i}{2}=\frac 12+\frac{\sqrt{3}}{2} i=e^{i \tfrac \pi 3}$

$\boxed{y=2 e^{i \tfrac \pi 3}}$

$t=1+i \implies |t|=\sqrt{1+1}=\sqrt{2}$
$\frac{t}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sq rt{2}} i=e^{i \tfrac \pi 4}$

$t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{\color{red}3}=2 \sqrt{2} e^{i \tfrac{{\color{red}3}\pi} 4}}$

$z=\frac{2 e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}}$

$z=\frac{1}{\sqrt{2}} \cdot \frac{e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{e^{i \tfrac{3\pi} 4}}$

$\frac{1}{\sqrt{2}}$ is the modulus. Now, use basic exponent rules to find the argument ;)
• October 2nd 2008, 05:16 PM
Dr Zoidburg
Moo, I just noticed that you wrote the equation as
$z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}$
you missed out the ^3 power up top:
$z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$

• October 3rd 2008, 09:56 AM
Moo
Quote:

Originally Posted by Dr Zoidburg
Moo, I just noticed that you wrote the equation as
$z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}$
you missed out the ^3 power up top:
$z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}$

Of course it changes it !
But the reasoning is completely the same. Please try to do it ;)
• October 3rd 2008, 10:00 AM
Dr Zoidburg
I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...
• October 3rd 2008, 10:04 AM
Moo
Quote:

Originally Posted by Dr Zoidburg
I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...

Oh sorry, I didn't see that...

Well, I don't see any mistake, after checking twice. However, given my state, don't take my word for granted at 100%.
• October 3rd 2008, 10:05 AM
Dr Zoidburg
I'll just hope it's either correct or the marker's in the same state you're in when s/he marks it!
thanks for the help