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Solve two quadratic equations (with real coefficients) in the complex numbers. The solutions have non-zero imaginary part, fractions can appear (but the denominators are rather small).

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Lösen Sie die folgenden quadratischen Gleichungen in den komplexen Zahlen. Die Gleichungen sind so ausgewählt, dass Sie die Lösungen in der Form $a+bi$ für Bruchzahlen (bzw. ganze Zahlen) $a$ und $b$ angeben können. Geben Sie Ihre Antwort ohne Klammern ein.

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Sie können diese quadratischen Gleichungen wie Gleichungen über den reellen Zahlen behandeln, nur dass Sie die Wurzel aus einer negativen Zahl ziehen müssen. Das ist in den komplexen Zahlen möglich: $\\sqrt{-a} = i\\sqrt{a}$ (für $a\\in\\mathbb R_{> 0}$).

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Im ersten Fall schreiben Sie die Gleichung um als $\\simplify[fractionNumbers]{(x+{p/2})^2 - {abs(p)/2}^2 + {q}}$, also sind die Lösungen $\\var[fractionnumbers]{-p/2} \\pm i\\sqrt{\\var[fractionNumbers]{-(p/2)^2+q}}$, also $\\var[fractionNumbers]{c1}$ und $\\var[fractionNumbers]{c2}$.

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Im zweiten Fall schreiben Sie die Gleichung um als $\\simplify[fractionNumbers]{(x+{pb/2})^2 - {abs(pb)/2}^2 + {qb}}$, also sind die Lösungen $\\var[fractionnumbers]{-pb/2} \\pm i\\sqrt{\\var[fractionNumbers]{-(pb/2)^2+qb}}$, also $\\var[fractionNumbers]{c1b}$ und $\\var[fractionNumbers]{c2b}$.

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Gegeben ist die Gleichung $\\simplify[all, fractionNumbers]{x^2 + {p}*x + {q}} = 0$.

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Die Lösungen sind [[0]] und [[1]].

\n

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Gegeben ist die Gleichung $\\simplify[all, fractionNumbers]{x^2 + {pb}*x + {qb}} = 0$.

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Die Lösungen sind [[0]] und [[1]].

\n

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