Q.The number $\dfrac{(1-i)^3}{1-i^3}$ is equal to _____.

The multiplicative inverse of √5+3i is (√5−3i)/14.

For a complex number z = a+bi, its multiplicative inverse is $\dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-bi}{a^2+b^2}$.

Here a=√5, b=3, so $a^2+b^2 = 5+9=14$.

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$i^{-35} = i$, i.e., in the form a+ib this is 0+1i.

Powers of i cycle with period 4: $i^1=i, i^2=-1, i^3=-i, i^4=1$, and this pattern repeats.

$i^{-35} = \dfrac{1}{i^{35}}$. Since $35 = 4\times8+3$, $i^{35}=i^{3}=-i$.

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The multiplicative inverse of a nonzero complex number $z$ is $\dfrac{\bar z}{|z|^2}$.

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$1+i+i^2+i^3+i^4 = 1$.

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$\left(\dfrac13+3i\right)^3 = -\dfrac{242}{27}-26i$.

Use $(a+bi)^3 = (a^3-3ab^2) + i(3a^2b - b^3)$ with $a=\dfrac13$, $b=3$.

Real part: $a^3-3ab^2 = \left(\dfrac13\right)^3 - 3\cdot\dfrac13\cdot3^2 = \dfrac{1}{27} - 9 = \dfrac{1}{27}-\dfrac{243}{27} = -\dfrac{242}{27}$.

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Writing $i$ and $-i$ in polar form and taking their principal square roots, the imaginary parts cancel on addition, leaving $\sqrt{2}$.

Write $i$ in polar (trigonometric) form: $i = \cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}$.

Step 1: Square root of $i$.

By De Moivre's theorem, the principal square root is

$\sqrt{i} = \cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4} = \dfrac{\sqrt2}{2} + i\dfrac{\sqrt2}{2}$

Step 2: Square root of $-i$.

Write $-i = \cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)$, so its principal square root is …

Reducing exponents mod $4$ gives $i^9=i$ and $i^{19}=-i$, which cancel to give $0$.

Since $i^4=1$, powers of $i$ repeat every $4$ steps. Reduce each exponent modulo $4$:

$i^9 = i^{4\times2+1} = (i^4)^2\cdot i = 1\cdot i = i$

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Using the principal square roots of $i$ and $-i$ in polar form, the sum simplifies to $\sqrt{2}$.

$i$ has modulus $1$ and argument $\dfrac{\pi}{2}$, so its principal square root is

$\sqrt{i}=\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}(1+i).$

$-i$ has modulus $1$ and argument $-\dfrac{\pi}{2}$, so its principal square root is …

Rationalising $\dfrac{1}{3-4i}$ using the conjugate $3+4i$ gives $\dfrac{3}{25}+\dfrac{4}{25}i$.

The multiplicative inverse of $3-4i$ is $\dfrac{1}{3-4i}$. Multiply numerator and denominator by the conjugate $3+4i$:

$\dfrac{1}{3-4i}\times\dfrac{3+4i}{3+4i}=\dfrac{3+4i}{3^2+4^2}=\dfrac{3+4i}{9+16}=\dfrac{3+4i}{25}.$ …

Since $|1-i|=\sqrt2$, the modulus of its reciprocal is $\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$.

For any nonzero complex numbers, $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$.

Here $z_1=1$ (modulus $1$) and $z_2=1-i$, whose modulus is

$|1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}.$

So …

Multiply and divide by the conjugate to get the multiplicative inverse $\dfrac{\sqrt5-3i}{14}$.

For a complex number $z = a+ib$, its multiplicative inverse is

$z^{-1} = \dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-ib}{a^2+b^2}$

Here $z = \sqrt5 + 3i$, so $a=\sqrt5,\ b=3$.

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