If z=a+bi and z* is the conjugate of z. Find the value of a and b when 2/z + 1/z* = 1-i. ?

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If z=a+bi and z* is the conjugate of z. Find the value of a and b when 2/z + 1/z* = 1-i. ?

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Answer 1 · 2017-11-11T09:28:06

The answer is $a = \frac{3}{10}$ $a=3/10$ and $b = \frac{9}{10}$ $b=9/10$

Explanation:

I use $¯ z$ $barz$ for the conjugate.

We need

$i^{2} = - 1$ $i^2=-1$

$(a + i b) (a - i b) = a^{2} - (i^{2}) b^{2} = a^{2} + b^{2}$ $(a+ib)(a-ib)=a^2-(i^2)b^2=a^2+b^2$

The complex number is $z = a + i b$ $z=a+ib$

The conjugate is $¯ z = a - i b$ $barz=a-ib$

$\frac{2}{z} = \frac{2}{a + i b} = \frac{2 (a - i b)}{(a + i b) (a - i b)} = \frac{2 (a - i b)}{a^{2} + b^{2}}$ $2/z=2/(a+ib)=(2(a-ib))/((a+ib)(a-ib))=(2(a-ib))/(a^2+b^2)$

$\frac{1}{¯ z} = \frac{1}{a - i b} = \frac{a + i b}{(a - i b) (a + i b)} = \frac{a + i b}{a^{2} + b^{2}}$ $1/barz=1/(a-ib)=(a+ib)/((a-ib)(a+ib))=(a+ib)/(a^2+b^2)$

Therefore,

$\frac{2}{z} + \frac{1}{¯ z} = \frac{2 (a - i b)}{a^{2} + b^{2}} + \frac{a + i b}{a^{2} + b^{2}}$ $2/z+1/barz=(2(a-ib))/(a^2+b^2)+(a+ib)/(a^2+b^2)$

$= \frac{2 a - 2 i b + a + i b}{a^{2} + b^{2}}$ $=(2a-2ib+a+ib)/(a^2+b^2)$

$= \frac{3 a - i b}{a^{2} + b^{2}}$ $=(3a-ib)/(a^2+b^2)$

$= 1 - i$ $=1-i$

Comparing the real parts and the imaginary parts

$\frac{3 a}{a^{2} + b^{2}} = 1$ $(3a)/(a^2+b^2)=1$

$\frac{b}{a^{2} + b^{2}} = 1$ $b/(a^2+b^2)=1$

Therefore,

$3 a = b$ $3a=b$

$\frac{3 a}{a^{2} + {(3 a)}^{2}} = 1$ $(3a)/(a^2+(3a)^2)=1$

$3 a = 10 a^{2}$ $3a=10a^2$

$10 a^{2} - 3 a = 0$ $10a^2-3a=0$

$a (10 a - 3) = 0$ $a(10a-3)=0$

Therefore,

$a = 0$ $a=0$ or $a = \frac{3}{10}$ $a=3/10$

Reject $a = 0$ $a=0$

$a = \frac{3}{10}$ $a=3/10$ , $\Rightarrow$ $=>$ , $b = \frac{9}{10}$ $b=9/10$

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If z=a+bi and z* is the conjugate of z. Find the value of a and b when 2/z + 1/z* = 1-i. ?

1 Answer

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