Answer
Verified
425.1k+ views
Hint: First of all put $z=x+iy$ in $v=\dfrac{1-iz}{z-i}$. Then simplify the v by separating real and imaginary part of it then put $\left| v \right|=1$ to find the value of z.
Complete step-by-step answer:
Here we are given that $z=x+iy$ and $v=\dfrac{1-iz}{z-i}$, we have to show that if $\left| v \right|=1$, then z is purely real.
First of all, let us take given expression,
$v=\dfrac{1-iz}{z-i}$
Here we know that $z=x+iy$. By putting the value of z in the above expression, we get;
$\begin{align}
& v=\dfrac{1-i\left( x+iy \right)}{\left( x+iy \right)-i} \\
& or \\
& v=\dfrac{1-ix-{{\left( i \right)}^{2}}y}{x+iy-i} \\
\end{align}$
As we know that i is an imaginary number & $i=\sqrt{-1}$, therefore we get ${{i}^{2}}=-1$
By putting the value of ${{i}^{2}}$ in above expression, we get,
$\begin{align}
& v=\dfrac{1-ix-\left( -1 \right)y}{x+iy-i} \\
& or \\
& v=\dfrac{1-ix+y}{x+iy-i} \\
\end{align}$
By separating real terms and imaginary terms in numerator denominator, we get,
$v=\dfrac{\left( 1+y \right)-ix}{x+i\left( y-1 \right)}$
Now, to rationalise the denominator, we will multiply numerator and denominator of above expression by $x-i\left( y-1 \right)$, we get,
$v=\dfrac{\left[ \left( 1+y \right)-ix \right]}{\left[ x+i\left( y-1 \right) \right]}\times \dfrac{\left[ x-i\left( y-1 \right) \right]}{\left[ x-i\left( y-1 \right) \right]}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By applying this in denominator of above expression, we get,
$v=\dfrac{\left[ \left( 1+y \right)-ix \right]\times \left[ x-i\left( y-1 \right) \right]}{\left[ {{\left( x \right)}^{2}}+{{\left( i\left( y-1 \right) \right)}^{2}} \right]}$
By simplifying the above expression, we get,
$v=\dfrac{\left( 1+y \right).x-i\left( y-1 \right)\left( y+1 \right)-ix.x+\left( ix \right)\left( i\left( y-1 \right) \right)}{\left[ {{x}^{2}}-{{i}^{2}}{{\left( y-1 \right)}^{2}} \right]}$
By further simplifying above expression, we get,
$v=\dfrac{\left( x+yx \right)-i\left( {{y}^{2}}-1 \right)-i{{x}^{2}}+{{i}^{2}}\left( xy-x \right)}{\left[ {{x}^{2}}-{{i}^{2}}{{\left( y-1 \right)}^{2}} \right]}$
By putting the value of ${{i}^{2}}=1$ in above expression, we get,
$\begin{align}
& v=\dfrac{\left( x+yx \right)-\left( i \right)\left( {{y}^{2}}-1 \right)-i{{x}^{2}}+\left( -1 \right)\left( xy-x \right)}{\left[ {{x}^{2}}-\left( -1 \right){{\left( y-1 \right)}^{2}} \right]} \\
& v=\dfrac{\left[ \left( x+yx \right)-i\left( {{y}^{2}}-1 \right)-i{{x}^{2}}-xy+x \right]}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \\
\end{align}$
By further simplifying the above expression, we get
$v=\dfrac{2x+i\left( -{{y}^{2}}+1-{{x}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}$
By separating real and imaginary term of above expression, we get,
$v=\dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}+\dfrac{i\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}$
Now we know that if we have $M=P+iQ$, then $M=\sqrt{{{P}^{2}}+{{Q}^{2}}}$. Using this we get,
\[\left| v \right|=\sqrt{{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}}\]
Also, as we are given that $\left| v \right|=1$, therefore, we get
\[\left| v \right|=\sqrt{{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}}=1\]
By squaring both sides, we get
\[{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}=1\]
By simplifying the above equation, we get,
\[\dfrac{4{{x}^{2}}+{{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}^{2}}}{{{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}}}=1\]
By cross multiplying the above equation, we get,
\[\begin{align}
& 4{{x}^{2}}+{{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}^{2}}={{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}} \\
& or \\
& 4{{x}^{2}}+{{\left( 1-\left( {{x}^{2}}+{{y}^{2}} \right) \right)}^{2}}={{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}} \\
\end{align}\]
As we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ by applying this in above equation, we get,
\[\begin{align}
& 4{{x}^{2}}+\left[ 1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right) \right]={{\left[ {{x}^{2}}+{{y}^{2}}+1-2y \right]}^{2}} \\
& =4{{x}^{2}}+1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left[ \left( {{x}^{2}}+{{y}^{2}} \right)+\left( 1-2y \right) \right]}^{2}} \\
\end{align}\]
By applying ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$in RHS of above equation, we get,
\[=4{{x}^{2}}+1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}+{{\left( 1-2y \right)}^{2}}+2\left( 1-2y \right)\left( {{x}^{2}}+{{y}^{2}} \right)\]
By cancelling the like terms from LHS and RHS, we get,
\[=4{{x}^{2}}+1-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( 1-2y \right)}^{2}}+2\left( 1-2y \right)\left( {{x}^{2}}+{{y}^{2}} \right)\]
By simplifying above equation, we get,
\[\Rightarrow 4{{x}^{2}}+1-2{{x}^{2}}-2{{y}^{2}}=1+4{{y}^{2}}-4y+2\left( {{x}^{2}}+{{y}^{2}}-2{{x}^{2}}y-2{{y}^{3}} \right)\]
By rearranging the terms, we get,
\[\begin{align}
& \Rightarrow \left( 4{{x}^{2}}-2{{x}^{2}} \right)+1-2{{y}^{2}}=1+\left[ 4{{y}^{2}}+2{{y}^{2}} \right]+2{{x}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
& \Rightarrow 2{{x}^{2}}+1-2{{y}^{2}}=1+6{{y}^{2}}+2{{x}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
\end{align}\]
By cancelling the like terms, we get,
$\begin{align}
& \Rightarrow -2{{y}^{2}}=6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
& or\ 6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}+2{{y}^{2}}=0 \\
& or\ 8{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}=0 \\
\end{align}$
By taking – 4y common, we get,
$-4y\left[ -2y+1+{{x}^{2}}+{{y}^{2}} \right]=0$
We can also write it as;
$-4y\left[ \left( {{y}^{2}}-2y+1 \right)+{{x}^{2}} \right]=0$
Here we can write $\left( {{y}^{2}}-2y+1 \right)={{\left( y-1 \right)}^{2}}$, therefore we get,
$-4y\left[ {{\left( y-1 \right)}^{2}}+{{x}^{2}} \right]=0$
As we can see that $\left[ {{\left( y-1 \right)}^{2}}+{{x}^{2}} \right]$ cannot be zero as they are always positive due to square, therefore we get,
$\begin{align}
& -4y=0 \\
& or\ y=0 \\
\end{align}$
Now, we know that $z=x+iy$
By putting y = 0, we get,
z = x
As z does not contain iota or i and contain only real number x, therefore it is purely real.
Hence, we have shown that z is purely real if $\left| v \right|=1$ where $v=\dfrac{1-iz}{z-i}$.
Note: Students must cross check every equation while solving the question because students often leave one or other term and get the wrong answer. Also student can cross check their answer by putting the value of z = x in expression for v and check if $\left| v \right|=1$ or not.
Complete step-by-step answer:
Here we are given that $z=x+iy$ and $v=\dfrac{1-iz}{z-i}$, we have to show that if $\left| v \right|=1$, then z is purely real.
First of all, let us take given expression,
$v=\dfrac{1-iz}{z-i}$
Here we know that $z=x+iy$. By putting the value of z in the above expression, we get;
$\begin{align}
& v=\dfrac{1-i\left( x+iy \right)}{\left( x+iy \right)-i} \\
& or \\
& v=\dfrac{1-ix-{{\left( i \right)}^{2}}y}{x+iy-i} \\
\end{align}$
As we know that i is an imaginary number & $i=\sqrt{-1}$, therefore we get ${{i}^{2}}=-1$
By putting the value of ${{i}^{2}}$ in above expression, we get,
$\begin{align}
& v=\dfrac{1-ix-\left( -1 \right)y}{x+iy-i} \\
& or \\
& v=\dfrac{1-ix+y}{x+iy-i} \\
\end{align}$
By separating real terms and imaginary terms in numerator denominator, we get,
$v=\dfrac{\left( 1+y \right)-ix}{x+i\left( y-1 \right)}$
Now, to rationalise the denominator, we will multiply numerator and denominator of above expression by $x-i\left( y-1 \right)$, we get,
$v=\dfrac{\left[ \left( 1+y \right)-ix \right]}{\left[ x+i\left( y-1 \right) \right]}\times \dfrac{\left[ x-i\left( y-1 \right) \right]}{\left[ x-i\left( y-1 \right) \right]}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By applying this in denominator of above expression, we get,
$v=\dfrac{\left[ \left( 1+y \right)-ix \right]\times \left[ x-i\left( y-1 \right) \right]}{\left[ {{\left( x \right)}^{2}}+{{\left( i\left( y-1 \right) \right)}^{2}} \right]}$
By simplifying the above expression, we get,
$v=\dfrac{\left( 1+y \right).x-i\left( y-1 \right)\left( y+1 \right)-ix.x+\left( ix \right)\left( i\left( y-1 \right) \right)}{\left[ {{x}^{2}}-{{i}^{2}}{{\left( y-1 \right)}^{2}} \right]}$
By further simplifying above expression, we get,
$v=\dfrac{\left( x+yx \right)-i\left( {{y}^{2}}-1 \right)-i{{x}^{2}}+{{i}^{2}}\left( xy-x \right)}{\left[ {{x}^{2}}-{{i}^{2}}{{\left( y-1 \right)}^{2}} \right]}$
By putting the value of ${{i}^{2}}=1$ in above expression, we get,
$\begin{align}
& v=\dfrac{\left( x+yx \right)-\left( i \right)\left( {{y}^{2}}-1 \right)-i{{x}^{2}}+\left( -1 \right)\left( xy-x \right)}{\left[ {{x}^{2}}-\left( -1 \right){{\left( y-1 \right)}^{2}} \right]} \\
& v=\dfrac{\left[ \left( x+yx \right)-i\left( {{y}^{2}}-1 \right)-i{{x}^{2}}-xy+x \right]}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \\
\end{align}$
By further simplifying the above expression, we get
$v=\dfrac{2x+i\left( -{{y}^{2}}+1-{{x}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}$
By separating real and imaginary term of above expression, we get,
$v=\dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}+\dfrac{i\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}$
Now we know that if we have $M=P+iQ$, then $M=\sqrt{{{P}^{2}}+{{Q}^{2}}}$. Using this we get,
\[\left| v \right|=\sqrt{{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}}\]
Also, as we are given that $\left| v \right|=1$, therefore, we get
\[\left| v \right|=\sqrt{{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}}=1\]
By squaring both sides, we get
\[{{\left( \dfrac{2x}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}+{{\left( \dfrac{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]} \right)}^{2}}=1\]
By simplifying the above equation, we get,
\[\dfrac{4{{x}^{2}}+{{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}^{2}}}{{{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}}}=1\]
By cross multiplying the above equation, we get,
\[\begin{align}
& 4{{x}^{2}}+{{\left( 1-{{x}^{2}}-{{y}^{2}} \right)}^{2}}={{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}} \\
& or \\
& 4{{x}^{2}}+{{\left( 1-\left( {{x}^{2}}+{{y}^{2}} \right) \right)}^{2}}={{\left[ {{x}^{2}}+{{\left( y-1 \right)}^{2}} \right]}^{2}} \\
\end{align}\]
As we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ by applying this in above equation, we get,
\[\begin{align}
& 4{{x}^{2}}+\left[ 1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right) \right]={{\left[ {{x}^{2}}+{{y}^{2}}+1-2y \right]}^{2}} \\
& =4{{x}^{2}}+1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left[ \left( {{x}^{2}}+{{y}^{2}} \right)+\left( 1-2y \right) \right]}^{2}} \\
\end{align}\]
By applying ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$in RHS of above equation, we get,
\[=4{{x}^{2}}+1+{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}+{{\left( 1-2y \right)}^{2}}+2\left( 1-2y \right)\left( {{x}^{2}}+{{y}^{2}} \right)\]
By cancelling the like terms from LHS and RHS, we get,
\[=4{{x}^{2}}+1-2\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( 1-2y \right)}^{2}}+2\left( 1-2y \right)\left( {{x}^{2}}+{{y}^{2}} \right)\]
By simplifying above equation, we get,
\[\Rightarrow 4{{x}^{2}}+1-2{{x}^{2}}-2{{y}^{2}}=1+4{{y}^{2}}-4y+2\left( {{x}^{2}}+{{y}^{2}}-2{{x}^{2}}y-2{{y}^{3}} \right)\]
By rearranging the terms, we get,
\[\begin{align}
& \Rightarrow \left( 4{{x}^{2}}-2{{x}^{2}} \right)+1-2{{y}^{2}}=1+\left[ 4{{y}^{2}}+2{{y}^{2}} \right]+2{{x}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
& \Rightarrow 2{{x}^{2}}+1-2{{y}^{2}}=1+6{{y}^{2}}+2{{x}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
\end{align}\]
By cancelling the like terms, we get,
$\begin{align}
& \Rightarrow -2{{y}^{2}}=6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\
& or\ 6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}+2{{y}^{2}}=0 \\
& or\ 8{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}=0 \\
\end{align}$
By taking – 4y common, we get,
$-4y\left[ -2y+1+{{x}^{2}}+{{y}^{2}} \right]=0$
We can also write it as;
$-4y\left[ \left( {{y}^{2}}-2y+1 \right)+{{x}^{2}} \right]=0$
Here we can write $\left( {{y}^{2}}-2y+1 \right)={{\left( y-1 \right)}^{2}}$, therefore we get,
$-4y\left[ {{\left( y-1 \right)}^{2}}+{{x}^{2}} \right]=0$
As we can see that $\left[ {{\left( y-1 \right)}^{2}}+{{x}^{2}} \right]$ cannot be zero as they are always positive due to square, therefore we get,
$\begin{align}
& -4y=0 \\
& or\ y=0 \\
\end{align}$
Now, we know that $z=x+iy$
By putting y = 0, we get,
z = x
As z does not contain iota or i and contain only real number x, therefore it is purely real.
Hence, we have shown that z is purely real if $\left| v \right|=1$ where $v=\dfrac{1-iz}{z-i}$.
Note: Students must cross check every equation while solving the question because students often leave one or other term and get the wrong answer. Also student can cross check their answer by putting the value of z = x in expression for v and check if $\left| v \right|=1$ or not.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE