# Thread: Complex No question 1

1. ## 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.

2. ## Re: Complex No question 1

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

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

b) For $\displaystyle z_1z_2$ to be purely imaginary number you must have
$\displaystyle \frac{\pi}{6}+\delta$ be an odd multiple of $\displaystyle \frac{\pi}{2}$.

3. ## 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!

4. ## Re: Complex No question 1

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

Suppose that $\displaystyle z=r\exp(i\theta)$ where $\displaystyle r\ge 0$ is a positive real number if and only if $\displaystyle \theta=2k\pi,~k\in\marhbb{Z}$.

5. ## Re: Complex No question 1

Is this a rule, thank you!

6. ## Re: Complex No question 1

Originally Posted by Tutu
Is this a rule, thank you!
I don't know what you mean by rule.

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

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

But in the question, we want a positive real number.

7. ## 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?

8. ## Re: Complex No question 1

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

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

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

Don't forget that you need $\displaystyle 0\le\delta\le 2\pi.$

9. ## 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!