If $$z=i(i+2)$$,then value of $$z4+4z3+6z2+4z$$ is

$$z=i(i+2)=$$−$$1+2i$$⇒$$(z+1)2=$$−$$2$$ ⇒$$z2+2z+3=0We$$ now $$dividez4+4z3+6z2+4z$$ by $$z2+2z+3z2+2z+3z4+4z3+6z2+4zz2+2z$$−$$1z4+2z3+3z22z3+3z2+4z2z3+4z2+6z$$∴ $$z4+4z3+6z2+4z=z2+2z+3z2+2z$$−$$1+3=0+3=3$$

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If $z = i (i + \sqrt{2})$ ,then value of $z^{4} + 4 z^{3} + 6 z^{2} + 4 z$ is

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Solution

$z = i (i + \sqrt{2}) = - 1 + \sqrt{2} i$
$\Rightarrow (z + 1)^{2} = - 2 \Rightarrow z^{2} + 2 z + 3 = 0$
We now divide
$z^{4} + 4 z^{3} + 6 z^{2} + 4 z by z^{2} + 2 z + 3$
$z^{2} + 2 z + 3 \sqrt{z^{4} + 4 z^{3} + 6 z^{2} + 4 z} [z^{2} + 2 z - 1$
$\frac{z^{4} + 2 z^{3} + 3 z^{2}}{2 z^{3} + 3 z^{2} + 4 z}$
$2 z^{3} + 4 z^{2} + 6 z$
$∴ z^{4} + 4 z^{3} + 6 z^{2} + 4 z$
$= (z^{2} + 2 z + 3) (z^{2} + 2 z - 1) + 3$
$= 0 + 3 = 3$

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Algebra of complex numbers

Remember concepts with our Masterclasses.

If z=i(i+2),then value of z4+4z3+6z2+4z is

z=i(i+2)=−1+2i⇒(z+1)2=−2 ⇒z2+2z+3=0We now dividez4+4z3+6z2+4z by z2+2z+3z2+2z+3z4+4z3+6z2+4zz2+2z−1z4+2z3+3z22z3+3z2+4z2z3+4z2+6z∴ z4+4z3+6z2+4z=z2+2z+3z2+2z−1+3=0+3=3

Ready to Test Your Skills?

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If $z = i (i + \sqrt{2})$ ,then value of $z^{4} + 4 z^{3} + 6 z^{2} + 4 z$ is