Let z = x + iy, then \(|z-1|=|z+1|\)
\(\Rightarrow \left| x+iy-1 \right| =\left| x+iy+1 \right| \)
\(\Rightarrow \left| (x-1)+iy \right| =\left| (x+1)-iy \right| \)
\(\Rightarrow\sqrt { (x-1)^{ 2 }+{ y }^{ 2 } } =\sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \)
\(\Rightarrow (x-1)^{ 2 }+{ y }^{ 2 }=(x+1)^{ 2 }+{ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+1-2x={ x }^{ 2 }+1+2x\Rightarrow 4x=0\Rightarrow x=0\)
\(\therefore \quad Re(z)=0\)

We have, \(z=\frac { (1+i)^{ 13 } }{ (1+i)^{ 7 } }\)
\(z=\frac { (1+i)^{ 13 } }{ (1+i)^{ 7 } } x\frac { (1+i{ ) }^{ 7 } }{ (1+i{ ) }^{ 7 } } \)
\([by\quad rationalising\quad the\quad denominator]\)
\(=\frac { (1+i{ ) }^{ 20 } }{ (1-{ i }^{ 2 }{ ) }^{ 7 } } =\frac { (1+i{ ) }^{ 20 } }{ (1+1{ ) }^{ 7 } } =\frac { (1+i{ ) }^{ 20 } }{ 2^{ 7 } } =\frac { 1 }{ 2^{ 7 } } (1+i{ ) }^{ 20 }\)
\( \frac { 1 }{ 2^{ 7 } } [(1+i{ )^{ 2 }] }^{ 10 }=\frac { 1 }{ 2^{ 7 } } (1+2i+{ i }^{ 2 }{ ) }^{ 10 }\)
\(\frac { 1 }{ 2^{ 7 } } (2i{ ) }^{ 10 }=\frac { (2{ ) }^{ 10 } }{ 2^{ 7 } } (i{ ) }^{ 10 }=(2{ ) }^{ 3 }({ i }^{ 2 })^{ 5 }=8(-1{ ) }^{ 5 }\)
\(z=-8\)
\(|z=\sqrt { (-8{ ) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } =8\)
and arg(z) = \(\pi \)  [z lies on negative X-axis]

\(\left| (1+i)(1+2i)(1+3i)....(1+ni) \right| =\left| x+iy \right| \)
\(\Rightarrow \left| 1+i \right| \left| 1+2i \right| \left| 1+3i \right| ....\left| 1+ni \right| =\left| x+iy \right| \)
\(\Rightarrow \sqrt { 1+1 } \sqrt { 1+4 } \sqrt { 1+9 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }\)
\( \Rightarrow \sqrt { 2 } \sqrt { 5 } \sqrt { 10 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)

Given \(\frac { \left( 1+i \right) x-2i }{ 3+i } +\frac { \left( 2-3i \right) y+i }{ 3-i } =i\)
\(\Rightarrow \frac { x+\left( x-2 \right) i }{ 3+i } +\frac { 2y+\left( 1-3y \right) i }{ 3-i } =i\)
\(\Rightarrow \frac { \left[ x+\left( x-2 \right) i \right] \left( 3-i \right) +\left[ 2y+\left( 1-3y \right) i \right] \left( 3+i \right) }{ \left( 3+i \right) \left( 3-i \right) } =i\)
\(\Rightarrow \left( 4x+9y-3 \right) +i\left( 2x-7y-3 \right) =10i\)
\(\Rightarrow 4x+9y-3=0\quad and\quad 2x-7y-3=10\)
\(Ans.\ x=3\ and \ y=-1\)

We have. \(x=-2-\sqrt { 3i } \)
\(\Rightarrow x=-2-\sqrt { 3i } \)
On squaring both sides, we get
\((x+2)^{2}=(-\sqrt{3} i)^{2} \Rightarrow x^{2}+4+4 x=3 i^{2}\)
\(\left[\because\left(z_{1}+z_{2}\right)^{2}=z_{1}^{2}+z_{2}^{2}+2 z_{1} z_{2}\right]\)
\(\Rightarrow \ x^{2}+4 x+4=-3 \quad\left[\because i^{2}=-1\right]\)
\(\Rightarrow \ x^{2}+4 x+7=0\)
Now divide \(2 x^{4}+5 x^{3}+7 x^{2}-x+41 \text { by } x^{2}+4 x+7\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 x^{2}-3 x+5 \\ x ^ { 2 } + 4 x + 7 \sqrt { 2 x ^ { 4 } + 5 x ^ { 3 } + 7 x ^ { 2 } - x + 4 1 } \)
\(\begin{aligned} 2 x^{4}+8 x^{3}+14 x^{2} \\ \frac{- \ -}{-3 x^{3}-7 x^{2}-x+41} \end{aligned}\)
\(\begin{aligned} 3 x^{3}-12 x^{2}-21 x^{} \\ \frac{+ \ + \ +}{5 x^{2}+20 x+35} \end{aligned}\)
\(\begin{aligned} {5 x^{2}+20 x+35} \\ \frac{- \ - \ -} {6} \end{aligned}\)
Thus, \({ 2x }^{ 4 }+{ 5x }^{ 3 }+{ 7x }^{ 2 }-x+41\)
\(=\left( { x }^{ 2 }+4x+7 \right) ({ 2x }^{ 2 }-3x+5)+6\)
\([\because dividend=quotient\times divisor+remainder]\)
\(=0\times (({ 2x }^{ 2 }-3x+5)+6=6\quad [\because { x }^{ 2 }+4x+7=0]\)