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

The solution to ((2+i)/(1-i)+(2-i)/(1+i)+i)^2 is 4+2i+((2+i)^2)/((1-i)^2)+((2-i)^2)/((1+i)^2)

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

simplify

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

4 + 2 i + \frac{( 2 + i ) ^{2}}{( 1 - i ) ^{2}} + \frac{( 2 - i ) ^{2}}{( 1 + i ) ^{2}}

Solution steps

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

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

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

Distribute parentheses

= \frac{2 + i}{1 - i} \cdot \frac{2 + i}{1 - i} + \frac{2 + i}{1 - i} \cdot \frac{2 - i}{1 + i} + \frac{2 + i}{1 - i} i + \frac{2 - i}{1 + i} \cdot \frac{2 + i}{1 - i} + \frac{2 - i}{1 + i} \cdot \frac{2 - i}{1 + i} + \frac{2 - i}{1 + i} i + i \frac{2 + i}{1 - i} + i \frac{2 - i}{1 + i} + ii

Simplify

\frac{2 + i}{1 - i} \cdot \frac{2 + i}{1 - i} + \frac{2 + i}{1 - i} \cdot \frac{2 - i}{1 + i} + \frac{2 + i}{1 - i} i + \frac{2 - i}{1 + i} \cdot \frac{2 + i}{1 - i} + \frac{2 - i}{1 + i} \cdot \frac{2 - i}{1 + i} + \frac{2 - i}{1 + i} i + i \frac{2 + i}{1 - i} + i \frac{2 - i}{1 + i} + ii : 2 i + \frac{( 2 + i ) ^{2}}{( 1 - i ) ^{2}} + \frac{( 2 - i ) ^{2}}{( 1 + i ) ^{2}} + 4

= 2 i + \frac{( 2 + i ) ^{2}}{( 1 - i ) ^{2}} + \frac{( 2 - i ) ^{2}}{( 1 + i ) ^{2}} + 4

Rewrite in standard complex form:

4 + 2 i + \frac{( 2 + i ) ^{2}}{( 1 - i ) ^{2}} + \frac{( 2 - i ) ^{2}}{( 1 + i ) ^{2}}

= 4 + 2 i + \frac{( 2 + i ) ^{2}}{( 1 - i ) ^{2}} + \frac{( 2 - i ) ^{2}}{( 1 + i ) ^{2}}

4 x^{2} - 9 x + 2 = 0

\frac{4 x ^{- 2} y ^{4}}{8 x ^{3} ( y ^{- 2} ) ^{3}}

7 x^{2} + 9 x y + 2 y^{2}

∣ x + 1 ∣ = ∣ 3 - x ∣

What is ((2+i)/(1-i)+(2-i)/(1+i)+i)^2 ?
The solution to ((2+i)/(1-i)+(2-i)/(1+i)+i)^2 is 4+2i+((2+i)^2)/((1-i)^2)+((2-i)^2)/((1+i)^2)