What is ((2+3i)(1-i)-(3+4i)^2)/(2i^{14)-i^{-7}} ?

The solution to ((2+3i)(1-i)-(3+4i)^2)/(2i^{14)-i^{-7}} is (12-23i)/(-2+1/i)

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simplify ((2+3i)(1-i)-(3+4i)^2)/(2i^{14)-i^{-7}}

simplify

\frac{( 2 + 3 i ) ( 1 - i ) - ( 3 + 4 i ) ^{2}}{2 i ^{14} - i ^{- 7}}

\frac{12 - 23 i}{- 2 + \frac{1}{i}}

Solution steps

\frac{( 2 + 3 i ) ( 1 - i ) - ( 3 + 4 i ) ^{2}}{2 i ^{14} - i ^{- 7}}

Simplify

(2 + 3 i) (1 - i) - (3 + 4 i)^{2} : - 23 i + 12

= \frac{- 23 i + 12}{2 i ^{14} - i ^{- 7}}

Simplify

2 i^{14} - i^{- 7} : - 2 + \frac{1}{i}

= \frac{- 23 i + 12}{- 2 + \frac{1}{i}}

Rewrite in standard complex form:

\frac{12 - 23 i}{- 2 + \frac{1}{i}}

= \frac{12 - 23 i}{- 2 + \frac{1}{i}}

25 > - 2 m

4 (u + 9) > 4 u + 2

∣ x - 6 ∣ < 5

(2 k - 9) (2 k - 1) < 0

What is ((2+3i)(1-i)-(3+4i)^2)/(2i^{14)-i^{-7}} ?
The solution to ((2+3i)(1-i)-(3+4i)^2)/(2i^{14)-i^{-7}} is (12-23i)/(-2+1/i)