Complex number z satisfying the equation z2 -1-i-7-17i- is-are

Z^2 ×(1+i)= 7+17i Let z=x+iy, where x,y ∈ℝ ⇒ (x+iy)^2= 7+17i(1+i) ⇒ x^2 y^2+2ixy = 7+17i(1+i)×(1 i)(1 i) ⇒ x^2 y^2+2ixy=10+24i2 ⇒ x^2 y^2=5, 2xy=12 ⇒ x^2 y^ ...

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Standard XII

Mathematics

Equality of 2 Complex Numbers

Complex numbe...

Question

Complex number $z$ satisfying the equation $z^{2} \times (1 + i) = - 7 + 17 i,$ is/are

z = - 3 + 2 i

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z = 3 + 2 i

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z = - 3 - 2 i

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z = 3 - 2 i

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Solution

The correct option is C $z = - 3 - 2 i$
$z^{2} \times (1 + i) = - 7 + 17 i$

Let $z = x + i y,$ where $x, y \in R$
$\Rightarrow (x + i y)^{2} = \frac{- 7 + 17 i}{(1 + i)} \Rightarrow x^{2} - y^{2} + 2 i x y = \frac{- 7 + 17 i}{(1 + i)} \times \frac{(1 - i)}{(1 - i)} \Rightarrow x^{2} - y^{2} + 2 i x y = \frac{10 + 24 i}{2} \Rightarrow x^{2} - y^{2} = 5, 2 x y = 12 \Rightarrow x^{2} - y^{2} = 5, x y = 6$

Solving the equations, we get
$x = 3, y = 2 or x = - 3, y = - 2$
Hence, $z = 3 + 2 i or z = - 3 - 2 i$

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Equality of 2 Complex Numbers

Standard XII Mathematics

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Complex number z satisfying the equation z2×(1+i)=−7+17i, is/are

The correct option is C z=−3−2iz2×(1+i)=−7+17i Let z=x+iy, where x,y∈R ⇒(x+iy)2=−7+17i(1+i)⇒x2−y2+2ixy=−7+17i(1+i)×(1−i)(1−i)⇒x2−y2+2ixy=10+24i2⇒x2−y2=5, 2xy=12⇒x2−y2=5, xy=6 Solving the equations, we get x=3,y=2 or x=−3,y=−2 Hence, z=3+2i or z=−3−2i

Complex number $z$ satisfying the equation $z^{2} \times (1 + i) = - 7 + 17 i,$ is/are