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I have to solve $$3z + 17i = iz + 11$$ and write $z$ in the form $a+bi$.

I substituted $a+bi$ for $z$ and got two simultaneous equations. $$3a+b = 11,\quad a-3b = -17$$ But am not getting the same answer. The answer is $z=5-4i$

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    $\begingroup$ In your second equation, the $-17$ should be $17$. $\endgroup$ – John Hughes Dec 7 '16 at 12:00
  • $\begingroup$ Oh my bad. Didn't realised my mistake $\endgroup$ – Ashalley Samuel Dec 7 '16 at 12:11
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$$3z + 17i = iz + 11$$ $$(3-i)z= 11- 17i$$ $$z=\frac{11-17i}{3-i}$$ Now multiply the numerator and denomerator by the conjugate of the denomaretor and separate the real and imaginary part.

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  • $\begingroup$ Thanks ...i mishandled my negative and positive signs. $\endgroup$ – Ashalley Samuel Dec 7 '16 at 12:12
  • $\begingroup$ You're welcome! $\endgroup$ – msm Dec 7 '16 at 12:13
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$$3z+17i=iz+11$$

substituting $z=a+ib$,

$$ 3(a+ib)+17i=i(a+ib)+11$$

$$ 3a +3bi+17i=ai-b+11$$

$$3a+i(3b+17)=(11-b)+ai$$

Equating real part to real and imaginary part to imaginary part.

$3a=11-b$ and $a=3b+17$. Now you have two equations and two variables. I shall let you conclude now.

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  • $\begingroup$ Thanks buddy. Well noted $\endgroup$ – Ashalley Samuel Dec 7 '16 at 12:14
  • $\begingroup$ @AshalleySamuel Always welcome.. $\endgroup$ – Vidyanshu Mishra Dec 7 '16 at 12:19
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You just need to solve the system of equations

$$3a+b=11$$ $$a-3b=17$$

Using elimination techniques,

$$3a+b=11$$ $$3a-9b=51$$

Which implies

$$10b=-40$$

Can you go from there?

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    $\begingroup$ I had a -17 which is where the problem started from. Thanks buddy $\endgroup$ – Ashalley Samuel Dec 7 '16 at 12:13
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$$3z+17i=iz+11 \Rightarrow z(3-i)=11-17i$$

$$ z=\frac{11-17i}{3-i}=\frac{(11-17i)(3+i)}{(3-i)(3+i)}=\frac{50-40i}{10}=5-4i$$

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