Question
If $a$ and $b$ are positive numbers such that $\frac{a z+0}{a z-b}$ is pure imaginary, then $|z|$ is equal to(a) 1(b) $\frac{a}{b}$(c) $\frac{b}{a}$(d) $\frac{a+b}{a-b}$
Step 1
Step 1: Given that $\frac{a z}{a z-b}$ is pure imaginary, we can write this as $\frac{a z+b}{a z-1} = k i$, where $k$ is a real number. Show more…
Show all steps
Your feedback will help us improve your experience
Uma Kumari and 62 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If $z=\frac{a+j b}{c+j d}$, where $a, b, c$ and $d$ are real quantities, show that (a) if $z$ is real then $\frac{a}{b}=\frac{c}{d}$ and (b) if $z$ is entirely imaginary then $\frac{a}{b}=-\frac{d}{c}$.
Complex numbers 1
Further problems
If $a>0$ and $z|z|+a z+2 a=0$ then $z$ must be (a) purely imaginary (b) a positive real number (c) a negative real number (d) 0
If $z=\frac{u+j d}{c+j d}$, where $a, b, c$ and $d$ are real quantities, show that (a) if $z$ is real then $\frac{a}{b}=\frac{c}{d}$ and (b) if $z$ is entirely imaginary then $\frac{a}{b}=-\frac{d}{c}$.
Further problems F.1
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD