Let $z_1 = 2 e^{i\pi/6}$ and $z_2 = re^{i\theta}$, where $r>0$ and $0\le\theta<2\pi$. Find the range of values of $r$ and $\theta$ for which $z_1z_2$ is:
a) a real number greater than $5$
b) a purely imaginary number with modulus less than $1$
This question was in my math textbook and I can't figure it out. There are no worked example to show the method in the textbook. Any hint on how I should approach it. I converted $z_1$ from polar to rectangular form and got $z_1 = \sqrt 3 + i$
I think $\theta$ has to be greater than $\pi$ and less than $2\pi$ because that will make value of $\sin\theta$ negative which would give the multiplication of $z_1z_2$ the form of $(a+b)(a-b)$ removing the imaginary part and leaving only real number. But I don't know how to get the number to be greater than $5$.
Answer in the textbook is
a) $r>5/2$ and $\theta=11\pi/6$
b) $r<1/2$ and $r > 0$ and $\theta=\pi/3$ or $\theta=4\pi/3$
Let $z_1=2e^{i\pi/6}$ and $z_2=re^{i\theta}$ where anyways $r\gt0$ and $0\leq\theta\lt2\pi$
Now the product $z_1z_2=2re^{i(\theta+i\pi/6)}$ The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>\frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2\pi$ which means $$ \theta+\pi/6=2\pi$$ giving $$\theta=11\pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so $$2r\lt1$$ Giving $$r\lt\frac{1}{2}$$ Also the amplitude is a positive quantity, $r\gt0$ Combining you get $$0\lt r\lt\frac{1}{2}$$ And to make the number purely imaginary the argument should be either $\pi/2$ or $3\pi/2$ $$\theta+\pi/6=\pi/2\space or\space 3\pi/2$$ So the values for $\theta$ will be $$\theta = \pi/3 \space or \space 4\pi/3$$ Hope this helps ….