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Question: Let \(z=x+iy\) and \(v=\dfrac{1-iz}{z-i}\), show that if \(\left| v \right|=1\), then z is purely re...

Let z=x+iyz=x+iy and v=1−izz−iv=\dfrac{1-iz}{z-i}, show that if ∣v∣=1\left| v \right|=1, then z is purely real.

Explanation

Solution

Hint: First of all put z=x+iyz=x+iy in v=1−izz−iv=\dfrac{1-iz}{z-i}. Then simplify the v by separating real and imaginary part of it then put ∣v∣=1\left| v \right|=1 to find the value of z.

Complete step-by-step answer:
Here we are given that z=x+iyz=x+iy and v=1−izz−iv=\dfrac{1-iz}{z-i}, we have to show that if ∣v∣=1\left| v \right|=1, then z is purely real.
First of all, let us take given expression,
v=1−izz−iv=\dfrac{1-iz}{z-i}
Here we know that z=x+iyz=x+iy. By putting the value of z in the above expression, we get;
v=1−i(x+iy)(x+iy)−i or v=1−ix−(i)2yx+iy−i \begin{aligned} & 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{aligned}
As we know that i is an imaginary number & i=−1i=\sqrt{-1}, therefore we get i2=−1{{i}^{2}}=-1
By putting the value of i2{{i}^{2}} in above expression, we get,
v=1−ix−(−1)yx+iy−i or v=1−ix+yx+iy−i \begin{aligned} & v=\dfrac{1-ix-\left( -1 \right)y}{x+iy-i} \\\ & or \\\ & v=\dfrac{1-ix+y}{x+iy-i} \\\ \end{aligned}
By separating real terms and imaginary terms in numerator denominator, we get,
v=(1+y)−ixx+i(y−1)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(y−1)x-i\left( y-1 \right), we get,
v=[(1+y)−ix][x+i(y−1)]×[x−i(y−1)][x−i(y−1)]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 (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}. By applying this in denominator of above expression, we get,
v=[(1+y)−ix]×[x−i(y−1)][(x)2+(i(y−1))2]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=(1+y).x−i(y−1)(y+1)−ix.x+(ix)(i(y−1))[x2−i2(y−1)2]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=(x+yx)−i(y2−1)−ix2+i2(xy−x)[x2−i2(y−1)2]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 i2=1{{i}^{2}}=1 in above expression, we get,
v=(x+yx)−(i)(y2−1)−ix2+(−1)(xy−x)[x2−(−1)(y−1)2] v=[(x+yx)−i(y2−1)−ix2−xy+x][x2+(y−1)2] \begin{aligned} & 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{aligned}
By further simplifying the above expression, we get
v=2x+i(−y2+1−x2)[x2+(y−1)2]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=2x[x2+(y−1)2]+i(1−x2−y2)[x2+(y−1)2]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+iQM=P+iQ, then M=P2+Q2M=\sqrt{{{P}^{2}}+{{Q}^{2}}}. Using this we get,
∣v∣=(2x[x2+(y−1)2])2+((1−x2−y2)[x2+(y−1)2])2\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 ∣v∣=1\left| v \right|=1, therefore, we get
∣v∣=(2x[x2+(y−1)2])2+((1−x2−y2)[x2+(y−1)2])2=1\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
(2x[x2+(y−1)2])2+((1−x2−y2)[x2+(y−1)2])2=1{{\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,
4x2+(1−x2−y2)2[x2+(y−1)2]2=1\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,

& 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{aligned}$$ As we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ by applying this in above equation, we get, $$\begin{aligned} & 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{aligned}$$ By applying ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$in RHS of above equation, we get, $$=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{aligned} & \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{aligned}$$ By cancelling the like terms, we get, $\begin{aligned} & \Rightarrow -2{{y}^{2}}=6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}} \\\ & or\ 6{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}+2{{y}^{2}}=0 \\\ & or\ 8{{y}^{2}}-4y-4{{x}^{2}}y-4{{y}^{3}}=0 \\\ \end{aligned}$ 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, $\begin{aligned} & -4y=0 \\\ & or\ y=0 \\\ \end{aligned}$ 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.