Complex number-binomial theorem, real and imaginary

**kingkaisai2** · July 31st 2006, 01:27 AM

If z=a+bi, where a and b are real, use the binomial theorem to find the real and imaginary parts of z^5 and (z*)^5, (z*) stand for conjugate of z

**Soroban** · July 31st 2006, 04:02 AM

Hello, kingkaisai2!

Exactly where is your difficulty?
. . You don't know the binomial theorem?
. . You can't handle $i^3$ ?
. . You just want to check your answer?
. . (It would be considerate to show us your answer.)

If $z=a+bi$ , where a and b are real, use the binomial theorem
to find the real and imaginary parts of $z^5$ and $(\overline{z})^5$

$(a + bi)^5\;=\;a^5 + 5a^4(bi) + 10a^3(bi)^2 +$ $10a^2(bi)^3 + 5a(bi)^4 + (bi)^5$

. . . . . . $= \;a^5 + 5a^4bi - 10a^3b^2 - 10a^2b^3i + 5ab^4 - b^5i$

. . . . . . $= \;\underbrace{(a^5 - 10a^3b + 5ab^4)} + \underbrace{(5a^4b - 10a^2b^3 - b^5)}i$
. . . . . . . . . . . . . .R . . . . . . . . . . . . . I

$(a - bi)^5\;=\;a^5 + 5a^4(-bi) + 10a^3(-bi)^2 +$ $10a^2(-bi)^3 + 5a(-bi)^4 + (-bi)^5$

. . . . . . $= \;a^5 - 5a^4bi - 10a^3b^2 + 10a^2b^3i + 5ab^4 + b^5i$

. . . . . . $= \;\underbrace{(a^5 - 10a^3b + 5ab^4)} - \underbrace{(5a^4b - 10a^2b^3 - b^5)}i$
. . . . . . . . . . . . . .R . . . . . . . . . . . . . I

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