10+i over 4-i
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Originally Posted by DRF4590 10+i over 4-i First do this: $\displaystyle \frac{10+i}{4-i}=\frac{(10+i)(4+i)}{(4-i)(4+i)}=\frac{40+14i+i^2}{16-i^2}$ Then simplify: $\displaystyle \frac{40+14i-1}{16-(-1)}$ Finally: $\displaystyle \frac{14i+39}{17}$ But check my arithmetic...
Originally Posted by DRF4590 10+i over 4-i Code: 10+i 4+i ----- * ------ 4 - i 4+i = (10*4+i*i)+(10*i+4*i) -------------------- 4*4 - i*i = (40-1)+i(14) ----------- 16+1 = 39+14i ------ 17 = (39/17) + (14/17) i -R.Mangalesh
10+i 4+i ----- * ------ 4 - i 4+i = (10*4+i*i)+(10*i+4*i) -------------------- 4*4 - i*i = (40-1)+i(14) ----------- 16+1 = 39+14i ------ 17 = (39/17) + (14/17) i
Last edited by CaptainBlack; Jan 24th 2007 at 04:53 AM. Reason: to tidy up formatting
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