Re: Complex No question 1

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**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 $\displaystyle \frac{\pi}{2}$.

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!

Re: Complex No question 1

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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 $\displaystyle \theta=2k\pi,~k\in\marhbb{Z}$.

Re: Complex No question 1

Is this a rule, thank you!

Re: Complex No question 1

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**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.

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?

Re: Complex No question 1

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

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!