Question

If $i{z}^3+z^2-z+i=0,$ then show that $\left|z\right|=1.$

Or

Find the real values of x and y, if $\frac{x-1}{3+i}+\frac{y-1}{3-i}=i.$

Answer 1

Given, $i{z}^3+z^2-z+i=0$

$\Rightarrow z^3+\frac{1}{i}{z}^2-\frac{1}{i}z+1=0$ [dividing both sides by i]

$\Rightarrow z^3-i{z}^2+iz+1=0[\because \frac{1}{i}=\frac{1}{i}\times \frac{i}{i}=-i]$

$\Rightarrow z^2(z-i)+i(z-i)=0[\because i^2=-1]$

$\Rightarrow (z^2+i)(z-i)=0$

$\Rightarrow z^2=-iorz=i$

Now, $z=i\Rightarrow \left|z\right|=\left|i\right|=1$

and $z^2=-1$

$\Rightarrow \left|z^2\right|=|-i|=1\Rightarrow {\left|z\right|}^2=1.$ $[\because \left|z^n\right|={\left|z\right|}^n]$

$\Rightarrow \left|z\right|=1$

Hence, in either case, we have $z^2=-1$ Hence, proved.

Or

Given, $\frac{x-1}{3+1}+\frac{y-1}{3-i}=i$

$\Rightarrow \frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}=i$

$\Rightarrow \frac{3x-ix-3+i+3y+iy-3-i}{9-i^2}=i$

$\Rightarrow \frac{(3x+3y-6)+i(y-x)}{9+1}=i$

$\Rightarrow (3x+3y-6)+i(y-x)=10i$

On equating the real and imaginary parts, we get

$3x+3y-6=0$ and $y-x=10$

$\Rightarrow x+y=2$ and $y-x=10$

On solving these equations, we get

$x=-4,y=6$