# Complex No question 1

• Nov 17th 2012, 07:39 AM
Tutu
Complex No question 1
Hi I need help for the following question,

1.) Let Z1 = 2cis(pi/6) and z2 = rcis(delta) where r>0, and 0<=delta<2pi.
FInd the range of values of r and delta for which z1z2 is
a.) a real number greater than 5
For the modulus of z1z2, I got r>5/2, which was right.
However for the delta, I'm not sure how I'm supposed to do it, it is (pi/6) + delta >5 and then solve the inequality? Seems wrong..
b.) a purely imaginary number with modulus less than 1
How do I find the imaginary number?

Thank you so much for your time, I really appreciate it!
J.
• Nov 17th 2012, 08:02 AM
Plato
Re: Complex No question 1
Quote:

Originally Posted by Tutu
1.) Let Z1 = 2cis(pi/6) and z2 = rcis(delta) where r>0, and 0<=delta<2pi. FInd the range of values of r and delta for which z1z2 is
a.) a real number greater than 5
For the modulus of z1z2, I got r>5/2, which was right.
However for the delta,
b.) a purely imaginary number with modulus less than 1
How do I find the imaginary number?

For $z_1z_2$ to be a real number greater than 5 you must have
$\frac{\pi}{6}+\delta=2k\pi$.

But also $0\le\delta\le 2\pi.$

b) For $z_1z_2$ to be purely imaginary number you must have
$\frac{\pi}{6}+\delta$ be an odd multiple of $\frac{\pi}{2}$.
• Nov 17th 2012, 08:39 AM
Tutu
Re: Complex No question 1
Thanks! May I know how you arrived at pi/6 + thetha = 2kpi?

Why 2kpi, to be exact.

Thank you so much!
• Nov 17th 2012, 08:47 AM
Plato
Re: Complex No question 1
Quote:

Originally Posted by Tutu
Thanks! May I know how you arrived at pi/6 + thetha = 2kpi? Why 2kpi, to be exact.

Suppose that $z=r\exp(i\theta)$ where $r\ge 0$ is a positive real number if and only if $\theta=2k\pi,~k\in\marhbb{Z}$.
• Nov 17th 2012, 09:03 AM
Tutu
Re: Complex No question 1
Is this a rule, thank you!
• Nov 17th 2012, 09:20 AM
Plato
Re: Complex No question 1
Quote:

Originally Posted by Tutu
Is this a rule, thank you!

I don't know what you mean by rule.

If $z=a+bi$ and $z\in\mathbb{R}$ it is necessary that $b=0$.

If $r\cos(\theta)+ir\sin(\theta)\in\mathbb{R}$ it is necessary that $\cos(\theta)=0$.
Thus $\theta=k\pi, k\in\mathbb{Z}$.

But in the question, we want a positive real number.
• Nov 17th 2012, 09:30 AM
Tutu
Re: Complex No question 1
But why is it that when you want a positive real number, it has to be 2kpi?
I think I kind of understand..
Does that meant that the angle has to be in the positive direction to get a positive real number hence 2kpi, not -2kpi or anything?
• Nov 17th 2012, 09:40 AM
Plato
Re: Complex No question 1
Quote:

Originally Posted by Tutu
But why is it that when you want a positive real number, it has to be 2kpi?
I think I kind of understand..
Does that meant that the angle has to be in the positive direction to get a positive real number hence 2kpi, not -2kpi or anything?

$\cos([2k+1]\pi)=-1$ so it must be an even multiple if $\pi$.

So in $\theta=2k\pi$, the integer $k$ can be positive or negative.

Recall $\cos(\theta)=\cos(-\theta)$.

Don't forget that you need $0\le\delta\le 2\pi.$
• Nov 17th 2012, 06:19 PM
Tutu
Re: Complex No question 1
How did you get
cos((2k+1)pi) = -1?

More specifically, why 2K+1 and how did you know the whole eqn equals to -1?

Thank you very much once again!