ISI-2016-02

Question

How many complex numbers $z$ are there such that $\mid z+1 \mid = \mid z+i \mid$ and $\mid z \mid = 5$ ?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Assuming $\color{violet}{z = x+ iy}$

Now, given that $\color{purple}{\mid z+1\mid = \mid z+i \mid }$

∴ $\color{purple}{\mid x+iy+1\mid = \mid x+iy+i \mid} $

Or, $\mid (x+1)+iy\mid = \mid (x)+(i+iy)\mid$

Or, $\sqrt{ (x+1)^2+(iy)^2} = \sqrt{ (x)^2+(i+iy)^2}$

Or, $\sqrt{ (x+1)^2+(y)^2} = \sqrt{ (x)^2+(i+iy)^2}$ $\qquad \qquad \big[∵\color{maroon}{i^2 =1}\big]$

Or, ${ (x+1)^2+(y)^2} ={ (x)^2+(i+iy)^2}$

Or, ${ x^2+2x+1+y^2} ={ x^2+ i^2+2.i.iy+i^2y^2}$

Or, ${ x^2+2x+1+y^2} ={ x^2+ i^2+2.i^2y+i^2y^2}$

Or, ${ x^2+2x+1+y^2} ={ x^2+ 1+2y+y^2}$

Or, $2x =2y$

Or, $\color{purple}{x=y}$

Also, given

$\color{violet}{\mid z \mid = 5}$

∴ $\sqrt{z^2} = 5$

Or,$ \sqrt{(x)^2+(iy)^2} = 5$

Or, $(x)^2+(iy)^2 = 25$

Or, $x^2 + i^2y^2 = 25$

Or, $\color{purple}{x^2 + y^2 = 25}$

When, 

$\color{maroon}{x}$	$\color{maroon}{\dfrac{\sqrt{5}}{2}}$	$\color{maroon}{4}$	$\color{maroon}{3}$
$\color{maroon}{y}$	$\color{maroon}{\dfrac{\sqrt{5}}{2}}$	$\color{maroon}{3}$	$\color{maroon}{4}$

∴ $\color{green}{\text{Possible complex numbers will be}}$ $\color{red}{3}$ i.e.

1.  $\color{blue}{x+iy \Rightarrow \dfrac{\sqrt{5}}{2} + i.\dfrac{\sqrt{5}}{2}}$
2.  $\color{blue}{x+iy \Rightarrow 4 + i.3 = 4+3i}$
3. $\color{blue}{x+iy \Rightarrow 3 + i.4 = 3+4i}$

Answer 2

simplify,

Z=X+iY
3+4i, 4+3i, (5/root(2))+(5/root(2))i

(D) 3

Answer 3

let Z=x+iy be a complex number

$\left | x+iy+1 \right |=\left | x+iy+i \right |$

$\sqrt{(x+1)^{2}+y^{2}}=\sqrt{x^{2}+(y+1)^{2}}$

x=y

now $\left | x+iy \right |$=5

$x^{2}+y^{2}=25$

$x=y=\frac{5}{\sqrt{2}}$

∣z+1∣=∣z+i∣ and ∣z∣=5

hence  $x=y=\frac{5}{\sqrt{2}}$ is only solution option B