What is ((z-1-i)/((z+i)(z-i)))^2 ?

The solution to ((z-1-i)/((z+i)(z-i)))^2 is ((z-1-i)^2)/((z^2+1)^2)

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simplify ((z-1-i)/((z+i)(z-i)))^2

simplify

(\frac{z - 1 - i}{( z + i ) ( z - i )})^{2}

\frac{( z - 1 - i ) ^{2}}{( z ^{2} + 1 ) ^{2}}

Solution steps

(\frac{z - 1 - i}{( z + i ) ( z - i )})^{2}

\frac{z - 1 - i}{( z + i ) ( z - i )} = \frac{z - 1 - i}{z ^{2} + 1}

= (\frac{z - 1 - i}{z ^{2} + 1})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( z - 1 - i ) ^{2}}{( z ^{2} + 1 ) ^{2}}

\frac{( x - 4 ) d x}{x ^{3}}

y^{'} (4) (x^{3} + y^{3} = 65)

- 3 x^{2}

\frac{( 10 + r ) 30}{40 + r}

8 (y - c) y^{'} (t)

What is ((z-1-i)/((z+i)(z-i)))^2 ?
The solution to ((z-1-i)/((z+i)(z-i)))^2 is ((z-1-i)^2)/((z^2+1)^2)