Question

Let $\omega$ be a complex number such that $\left|\frac{\omega-i}{\omega+3i}\right|=1$ (where $i=\sqrt{-1}$) and $|\omega|=7$. Then the value of $\left|\omega+\frac{3}{2}i\right|$ equals to

• A. $\frac{\sqrt{201}}{2}$
• B. 7
• C. 5
• D. $\frac{\sqrt{193}}{2}$

Answer 1

Answer: (D)

$\frac{\sqrt{193}}{2}$

Answer 2

Let $\omega = x+iy$. The condition $\left|\frac{\omega-i}{\omega+3i}\right|=1$ implies $|\omega-i|^2 = |\omega+3i|^2$. $|x+i(y-1)|^2 = |x+i(y+3)|^2$ $x^2 + (y-1)^2 = x^2 + (y+3)^2$ $y^2 - 2y + 1 = y^2 + 6y + 9$ $-2y + 1 = 6y + 9$ $8y = -8$ $y = -1$.

The condition $|\omega|=7$ implies $|\omega|^2=49$. $x^2 + y^2 = 49$ Substituting $y=-1$: $x^2 + (-1)^2 = 49$ $x^2 + 1 = 49$ $x^2 = 48$.

We need to find $\left|\omega+\frac{3}{2}i\right|$. Substitute $\omega = x+iy = x-i$: $\left|x-i+\frac{3}{2}i\right| = \left|x+i\left(\frac{1}{2}\right)\right|$ This is equal to $\sqrt{x^2 + \left(\frac{1}{2}\right)^2}$. Substitute $x^2=48$: $\sqrt{48 + \frac{1}{4}} = \sqrt{\frac{192+1}{4}} = \sqrt{\frac{193}{4}} = \frac{\sqrt{193}}{2}$.