Question

Can i get quick answer​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Express (-5i) * (i/8) in the form of a + ib ?

Sᴏʟᴜᴛɪᴏɴ :-

→ (-5i) * (i/8)

→ {(-5) * (1/8)} * (i * i)

→ (-5/8) * (i)²

Putting value of (i)² = (-1) now, { i is iota here.}

→ (-5/8) * (-1)

→ (5/8)

So, in a + ib form ,

→ (5/8) + 0i (Ans.)

where :-

• (5/8) = Real Part.
• (0i) = Imaginary Part.

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Extra :-

→ Integral Powers of iota (i)

1. i =√-1
2. i2 = -1
3. i3 = -i
4. i4 = 1

So,

• i^(4n+1) = i,
• i^(4n+2) = -1
• i^(4n+3) = -i
• i^(4n+4) = i^(4n) = 1

In other words,

• i^n = (-1)^(n/2), if n is an even integer.
• i^n = (-1)^{(n-1)/2}.i, if is an odd integer.

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Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: {\sf{( - 5i) \times \bigg( \frac{1i}{8} \bigg) }}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ - 5i \times \frac{1i}{8} }}$

$: \implies{\sf{ - 5 \times \frac{1}{8} \times( i \times i )}}$

we know that i = √-1

$: \implies{\sf{ - 5 \times \frac{1}{8} \times( \sqrt{ - 1} \times \sqrt{ - 1} ) }}$

$: \implies{\sf{ - 5 \times \frac{1}{8} \times - 1 }}$

$: \implies{\sf{ \frac{5}{8} + 0i}}$

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• i is called iota whose square is -1.
• Therefore, i² = -1.
• Number i and its real multiple are called Imaginary numbers.