What is (1-(2+3i)^2(1-2i))/(2i^{77)-i^{726}} ?

The solution to (1-(2+3i)^2(1-2i))/(2i^{77)-i^{726}} is -62/5+i 14/5

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simplify (1-(2+3i)^2(1-2i))/(2i^{77)-i^{726}}

simplify

\frac{1 - ( 2 + 3 i ) ^{2} ( 1 - 2 i )}{2 i ^{77} - i ^{726}}

- \frac{62}{5} + i \frac{14}{5}

Solution steps

\frac{1 - ( 2 + 3 i ) ^{2} ( 1 - 2 i )}{2 i ^{77} - i ^{726}}

2 i^{77} - i^{726} = 2 i + 1

= \frac{1 - ( 2 + 3 i ) ^{2} ( 1 - 2 i )}{2 i + 1}

(2 + 3 i)^{2} (1 - 2 i) = 19 + 22 i

= \frac{1 - ( 19 + 22 i )}{2 i + 1}

Apply complex arithmetic rule:

\frac{a + bi}{c + d i} = \frac{( c - d i ) ( a + bi )}{( c - d i ) ( c + d i )} = \frac{( a c + b d ) + ( b c - a d ) i}{c ^{2} + d ^{2}}

a = - 18, b = - 22, c = 1, d = 2

= \frac{( - 18 \cdot 1 + ( - 22 ) \cdot 2 ) + ( - 22 \cdot 1 - ( - 18 ) \cdot 2 ) i}{1 ^{2} + 2 ^{2}}

Refine

= \frac{- 62 + 14 i}{5}

Rewrite

\frac{- 62 + 14 i}{5}

in standard complex form:

- \frac{62}{5} + \frac{14}{5} i

= - \frac{62}{5} + \frac{14}{5} i

\frac{3}{4} = \frac{x}{9}

\frac{4}{6} + \frac{1}{3}

lo g_{9} (81) = L

x^{5} y + x^{3} y^{5}

\frac{5}{4} t = - 20

What is (1-(2+3i)^2(1-2i))/(2i^{77)-i^{726}} ?
The solution to (1-(2+3i)^2(1-2i))/(2i^{77)-i^{726}} is -62/5+i 14/5