Question

Find the complex number $\mathcal{Z}$, if $|\mathcal{Z}+1|=\mathcal{Z}+2\left(1+i\right)$.​

Answer 1

$|\mathcal{Z}+1|=\mathcal{Z}+2\left(1+i\right)$

Let us assume $|\mathcal{Z}= a+ib$

So,

In the equation we have :-

$|\mathcal{Z}+1|=\mathcal{Z}+2\left(1+i\right) \\ \\ \\|a + ib+1|=a + ib+2\left(1+i\right) \\ \\ \\|(a+1) + ib|=(a + 2) + (b + 2)i$

We already know that :-

$|\mathcal{Z}| = \sqrt{ {a}^{2} + {b}^{2} }$

Thus we have the equation as :-

$\sqrt{ {(a + 1)}^{2} + {b}^{2} } =(a + 2) + (b + 2)i$

On comparing the real and imaginary part of LHS and RHS we get :-

$b + 2 = 0 \\ \\ \\ b = ( - 2)$

Also,

$\sqrt{ {(a + 1)}^{2} + {b}^{2} } = a + 2 \\ \\ \\ \sqrt{ {a}^{2} + 2a + 1 + {( - 2)}^{2} } = a + 2 \\ \\ \\ {a}^{2} + 2a + 1 + 4 = {(a +2)}^{2} \\ \\ \\ {a}^{2} + 2a + 5 = {a}^{2} + 4a + 4 \\ \\ \\4a - 2a = 5 - 4 = 1 \\ \\ \\2a = 1 \\ \\ \\a = \frac{1}{2}$

Hence,

We have the values of a = 1/2 and b=(-2)

Putting in the form a+ib to get the complex number z

So,

$\mathcal{Z} = a + ib \\ \\ \\\mathcal{Z} = \frac{1}{2} - 2i$

Answer 2

Answer:

hope it is helpful to you.