Integral Maths - Complex numbers - Test C1

Could someone please help me with questions 6 and 9.

6) Given that z = 3 + 4i / 2 - 3i, the complex number w which satisfies the equation zw = 1 is given by w = _ + _i
(Give your answers as exact whole numbers or fractions in their lowest terms in the form a/b)

9) Find the values of a and b (with a > 0) which satisfy (a + bi)² = 5 + 12i
a = _, and b = _

Thank you if you are able to help.

Could someone please help me with questions 6 and 9.

6) Given that z = 3 + 4i / 2 - 3i, the complex number w which satisfies the equation zw = 1 is given by w = _ + _i
(Give your answers as exact whole numbers or fractions in their lowest terms in the form a/b)

9) Find the values of a and b (with a > 0) which satisfy (a + bi)² = 5 + 12i
a = _, and b = _

Thank you if you are able to help.

Hi, I’ve not done this integral but I do take further maths. For number 6 my best guess would be that for zw to equal 1 then w has to be the reciprocal of z so w = (2-3i)/(3+4i). To get it in a better form you could then multiply both parts of the fraction by the conjugate of the denominator. This is just a guess and could be wrong but I hope this helps.

Hi, thanks so much, I'll try that now.
Yep, that was so helpful, got it right now, thanks again 😁

That’s okay I’ve recently been covering the year 13 part of complex numbers so it’s just good practice and recap. I’ll have a look at the other question to see if I know how to do that.

Oh right, yeah, I've just started year 12 so it's my first time learning about complex numbers. Thanks ☺️

Oh right, yeah, I've just started year 12 so it's my first time learning about complex numbers. Thanks ☺️

So what I’ve done for this is expand the (a + bi)² to be a² + 2abi - b² . Using this you can then equate coefficients so a² - b² = 5 and 2ab = 12. The a² - b² can be factorised to (a +b)(a - b). The factors of 5 is just 1 and 5 so set one factor equal to each( I tried both ways round). I then used these equations to calculate b using the simultaneous equations. You can then use b to find a which is positive. You’ll get two values for b ( one positive, one negative) substitute the values into the 2ab = 12 equation to figure out which one is right. Again this is just how I approached it I may be wrong 😂 If you want anything of this explained a bit more just let me know

Thanks so much, again that was really helpful and correct, thanks ☺️

I’m glad it was helpful. Feel free to message me if you want anymore help with further maths!

So what I’ve done for this is expand the (a + bi)² to be a² + 2abi - b² . Using this you can then equate coefficients so a² - b² = 5 and 2ab = 12. The a² - b² can be factorised to (a +b)(a - b). The factors of 5 is just 1 and 5 so set one factor equal to each( I tried both ways round). I then used these equations to calculate b using the simultaneous equations. You can then use b to find a which is positive. You’ll get two values for b ( one positive, one negative) substitute the values into the 2ab = 12 equation to figure out which one is right. Again this is just how I approached it I may be wrong 😂 If you want anything of this explained a bit more just let me know